Happy new year to u all

benbuks: ^ ok...m on bed nw..mayb i ‘d try dem 2maro...quit cheep sha.

if u tink they are cheap, den we 'll patiently wait 4 ur solution...

Laplacian:
if u tink they are cheap, den we 'll patiently wait 4 ur solution...

futurewise11: please kindly help in breaking this word problem down

@futurewise11 I tried sending a message to your email buh it didnt go tru. Kindly get back to me when u get this pls...here's my email addy okun2010@yahoo.com. Wld really apric8 it.

new year questions .
express in factorial form
n(n+1)(n+2). . .(n+k)

1^2 . 2^2 . 3^2 . . . n^2


find the term independent of x in. [x + 1/x]^2n


.

Pat £inst£in:

..leme try ur questions..nw..might get bk to u later...

@ laplacian. u cn also try n post ur solution o dont wait for me oo..

The coefficient of x^4 in the expansion of (1 - x + x^2)^4 is __

SOLUTION

Let: y = 1 - x
(y + x^2)^4
Expanding by binomial expansion yields
y^4 + 4y^3x^2 + 6y^2x^4 + 4yx^6 + x^8
From the above:
y^4 + 4y^3x^2 + 6y^2x^4 gives a coefficient of x^4

Separately, we have
y^4 = (1 - x)^4 = 1 - 4x + 6x^2 - 4x^3 + x^4
We have 1 as coefficient of x^4 here

4y^3x^2 = 4[(1 - x)^3x^2] = 4[(1 - 3x + 3x^2 - x^3)x^2] = 4(x^2 - 3x^3 + 3x^4 - x^5) = 4x^2 - 12x^3 + 12x^4 - 4x^5
We have 12 as coefficient of x^4 here

6y^2x^4 = 6[(1 - x)^2x^4] = 6[(1 - 2x + x^2)x^4] = 6x^4 - 12x^5 + 6x^6
We have 6 as coefficient of x^4 here

Adding all coefficients of x^4 yields: 1 + 12 + 6 = 19
Hence, coefficient of x^4 = 19


Thanks again @aysuccess99

Pat £inst£in:

Hmm, doc of nairaland maths clinic, solve my 4 questions na

If 'a' is the sum of odd terms and 'b' the sum of even terms in the expansion (x + a)^b , then find a^2 - b^2

The coefficient of x^4 in the expansion of (1 - x + x^2)^4 is __

Sum the series:
[5/(3 • 6)] + [(5 • 7)/(3 • 6 • 9)] + [(5 • 7 • 9)/(3 • 6 • 9 • 12)] + ...~
NB: the • means multiplication as in, 2 • 3 = 6

Prove that tan^-1(x+z) = tan^-1(x) + 2sin(z)sin(z) - [(2sin(z))^2sin^2(z)]/2 + [(2sin(z)^3sin^3(z)]/3 - ...
Where cot(z) = x

Thanks in anticipation...

Seriously now, I'm cool wit u

for ur no.1, am not really sure what u mean by "odd and even terms", but here's a suggestion;
1.) expansion of (x+a)^b should contain only "odd & even terms", so that, from the condition of the question,
a+b=(x+a)^b, then multiplying thru by a-b, we obtain;
a²-b²=(x+a)^b•(a-b)..

2.) this one is quiet 'cheap', given
f(x)=
(1 - x + x²)⁴. Take the fouth order derivative of the above expression, set x=0, and divide the result by 4!...
Simply put, obtain f^iv(0)/4!
In case u ain't satisfied with that, here is an elementary
ALTERNATIVE SOLUTION
(1 - x + x²)⁴=[1+(x-1)x]⁴=1+4(x-1)x+6[(x-1)x]²+4[(x-1)x]³+[(x-1)x]⁴, inspection shows that only the last three terms contains x⁴, and the coefficient can be readily obtained from there.
3.) let # denote pi, the product of an expression in d variable r, from r=1 to n, by inspection, the numerator of each term of ur series is the product of an A.P with fomula; #(3+2r), similar, the denominator is the product of the natural numbers and some power of three, with the formula; 3ⁿ⁺¹•(n+1)!, so the n^th term of ur sequence is;
T_n=[#(3+2r)]/[3ⁿ⁺¹•(n+1)!]
Recall that,
#(3+2r)=5•7•9•11•13•15...
=#[(2+2r)•(3+2r)]/#(2+2r)
=(4•5•6•7....)/(4•6•8•10•12...)

#(3+2r)=

[(3+2n)!/3!]/(2ⁿ•3•4•5•6...)
=[(3+2n)!/3!]/(2ⁿ/2•[(3n+2)/2]!)
so that,
T_n=[(3+2n)!/3!]/(2ⁿ/2•[(3+2n)/2]!)/[3ⁿ⁺¹•(n+1)!]

the solution becomes too complicatd to paste here afterward...i'll paste the final solution soon...

Pat £inst£in:
The coefficient of x^4 in the expansion of (1 - x + x^2)^4 is __

SOLUTION

Let: y = 1 - x
(y + x^2)^4
Expanding by binomial expansion yields
y^4 + 4y^3x^2 + 6y^2x^4 + 4yx^6 + x^8
From the above:
y^4 + 4y^3x^2 + 6y^2x^4 gives a coefficient of x^4

Separately, we have
y^4 = (1 - x)^4 = 1 - 4x + 6x^2 - 4x^3 + x^4
We have 1 as coefficient of x^4 here

4y^3x^2 = 4[(1 - x)^3x^2] = 4[(1 - 3x + 3x^2 - x^3)x^2] = 4(x^2 - 3x^3 + 3x^4 - x^5) = 4x^2 - 12x^3 + 12x^4 - 4x^5
We have 12 as coefficient of x^4 here

6y^2x^4 = 6[(1 - x)^2x^4] = 6[(1 - 2x + x^2)x^4] = 6x^4 - 12x^5 + 6x^6
We have 6 as coefficient of x^4 here

Adding all coefficients of x^4 yields: 1 + 12 + 6 = 19
Hence, coefficient of x^4 = 19


Thanks again @aysuccess99

^

Nice one, though the problem has already been solved by AYSUCCESS99. Why not finish the rest if you have the solutions? The reason why solutions to some questions are not posted quickly is because typing them out isn't all that easy, it's not because your questions are too difficult or "hard" for gurus here to solve.

For the benefit of everyone, if anyone has a solution to his/her problem, it's better they post them if the questions are not attended to quickly; that way, others would learn from from them, instead of recycling the problems over and over again! I'm saying this because a lot of people post questions and leave them unsolved if the the clinic's active members do not make efforts to attempt them. We are building this all thing to encourage mathematics lovers of today and future leaders, so let's do it with all our hearts and might; it's not about competition!

Thank y'all and happy new year!

Laplacian:
for ur no.1, am not really sure what u mean by "odd and even terms", but here's a suggestion;
1.) expansion of (x+a)^b should contain only "odd & even terms", so that, from the condition of the question,
a+b=(x+a)^b, then multiplying thru by a-b, we obtain;
a²-b²=(x+a)^b•(a-b)..

2.) this one is quiet 'cheap', given
f(x)=
(1 - x + x²)⁴. Take the fouth order derivative of the above expression, set x=0, and divide the result by 4!...
Simply put, obtain f^iv(0)/4!
In case u ain't satisfied with that, here is an elementary
ALTERNATIVE SOLUTION
(1 - x + x²)⁴=[1+(x-1)x]⁴=1+4(x-1)x+6[(x-1)x]²+4[(x-1)x]³+[(x-1)x]⁴, inspection shows that only the last three terms contains x⁴, and the coefficient can be readily obtained from there.
3.) let # denote pi, the product of an expression in d variable r, from r=1 to n, by inspection, the numerator of each term of ur series is the product of an A.P with fomula; #(3r+2), similar, the denominator is the product of a G.P with the formula; 3ⁿ⁺¹•(n+1)!, so the n^th term of ur sequence is;
T_n=[#(3r+2)]/[3ⁿ⁺¹•(n+1)!]
Recall that,
#(3r+2)=5•7•9•11•13•15...
=#[(3r+1)•(3r+2)]/#(3r+1)
=(4•5•6•7....)/(4•6•8•10•12...)
(i proceed for the case in which n is even, use n+1 for the index of 2 if n is odd)
#(3r+2)=

[(3n+2)!/3!]/(2ⁿ•3•4•5•6...)
=[(3n+2)!/3!]/(2ⁿ/2•[(3n+2)/2]!)
so that,
T_n=[(3n+2)!/3!]/(2ⁿ/2•[(3n+2)/2]!)/[3ⁿ⁺¹•(n+1)!]

the solution becomes too complicatd to paste here afterward...i'll paste the final solution soon...

Ok, cool stuff Laplacian!

doubleDx:

^

Nice one, though the problem has already been solved by AYSUCCESS99. Why not finish the rest if you have the solutions? The reason why solutions to some questions are not posted quickly is because typing them out isn't all that easy, it's not because your questions are too difficult or "hard" for gurus here to solve.

For the benefit of everyone, if anyone has a solution to his/her problem, it's better they post them if the questions are not attended to quickly; that way, others would learn from from them, instead recycling the problems over and over again! I'm saying this because a lot of people post questions and leave them unsolved if the the clinic's active members do not make efforts to attempt them. We are building this all thing to encourage mathematics lovers of today and future leaders, so let's do it with all our hearts and might; it's not about competition!

Thank y'all and happy new year!

Aysuccess99 used normal expansion which is tedious - not a good idea in exam situations, but still got it correctly. I used binomial expansion which is easier; I actually solved it TODAY.

As for the other questions, I am yet to solve them that's why I posted all 4 questions.

ONE LOVE

Pat £inst£in:

Aysuccess99 used normal expansion which is tedious - not a good idea in exam situations, but still got it correctly. I used binomial expansion which is easier; I actually solved it TODAY.

As for the other questions, I am yet to solve them that's why I posted all 4 questions.

ONE LOVE

Oh, okay. It's okay....

Laplacian:
for ur no.1, am not really sure what u mean by "odd and even terms", but here's a suggestion;
1.) expansion of (x+a)^b should contain only "odd & even terms", so that, from the condition of the question,
a+b=(x+a)^b, then multiplying thru by a-b, we obtain;
a²-b²=(x+a)^b•(a-b)..

2.) this one is quiet 'cheap', given
f(x)=
(1 - x + x²)⁴. Take the fouth order derivative of the above expression, set x=0, and divide the result by 4!...
Simply put, obtain f^iv(0)/4!
In case u ain't satisfied with that, here is an elementary
ALTERNATIVE SOLUTION
(1 - x + x²)⁴=[1+(x-1)x]⁴=1+4(x-1)x+6[(x-1)x]²+4[(x-1)x]³+[(x-1)x]⁴, inspection shows that only the last three terms contains x⁴, and the coefficient can be readily obtained from there.
3.) let # denote pi, the product of an expression in d variable r, from r=1 to n, by inspection, the numerator of each term of ur series is the product of an A.P with fomula; #(3r+2), similar, the denominator is the product of a G.P with the formula; 3ⁿ⁺¹•(n+1)!, so the n^th term of ur sequence is;
T_n=[#(3r+2)]/[3ⁿ⁺¹•(n+1)!]
Recall that,
#(3r+2)=5•7•9•11•13•15...
=#[(3r+1)•(3r+2)]/#(3r+1)
=(4•5•6•7....)/(4•6•8•10•12...)
(i proceed for the case in which n is even, use n+1 for the index of 2 if n is odd)
#(3r+2)=

[(3n+2)!/3!]/(2ⁿ•3•4•5•6...)
=[(3n+2)!/3!]/(2ⁿ/2•[(3n+2)/2]!)
so that,
T_n=[(3n+2)!/3!]/(2ⁿ/2•[(3n+2)/2]!)/[3ⁿ⁺¹•(n+1)!]

the solution becomes too complicatd to paste here afterward...i'll paste the final solution soon...

I'm clear with d no 1 and 2 solutions, but not very ok wit d 3rd - d numerator series formula
D formula I hav for d numerator is #(5+(n-1)2)

Have a look at dis:

Sum the series
1 + 3/4 + 3•5/4•8 + 3•5•7/4•8•12 + ...

SOLUTION

(1+x)^n = 1 + nx + [n(n-1)/2!]x^2 + [n(n-1)(n-2)/3!]x^3 + ... + x^n
We find nx = 3/4 ...........(1)
[n(n-1)/2!]x^2 = 3•5/4•8 .........(2)
From (1)
x = 3/4n .......(3)
Putting (3) into (2)

[n(n-1)/2](3/4n)^2 = 3•5/4•8
9n(n-1)/32n^2 = 15/32
3(n-1)/n = 5
(1-1/n) = 5/3
1/n = 1-5/3
1/n = -2/3

n = -3/2 ......................

From x = 3/4n = 3/4(1/n) = 3/4(-2/3)

x = -1/2 ......................

Hence the series is (1 + x)^n = (1 - 1/2)^(-3/2) = 1 + 3/4 + 3•5/4•8 + 3•5•7/4•8•12 + ...
Solving (1 - 1/2)^(-3/2) gives 2(2)^(1/2) meaning 2 * square root of 2

Thanks a lot bro

Pat £inst£in:

I'm clear with d no 1 and 2 solutions, but not very ok wit d 3rd - d numerator series formula
D formula I hav for d numerator is #(5+(n-1)2)

Have a look at dis:

Sum the series
1 + 3/4 + 3•5/4•8 + 3•5•7/4•8•12 + ...

SOLUTION

(1+x)^n = 1 + nx + [n(n-1)/2!]x^2 + [n(n-1)(n-2)/3!]x^3 + ... + x^n
We find nx = 3/4 ...........(1)
[n(n-1)/2!]x^2 = 3•5/4•8 .........(2)
From (1)
x = 3/4n .......(3)
Putting (3) into (2)

[n(n-1)/2](3/4n)^2 = 3•5/4•8
9n(n-1)/32n^2 = 15/32
3(n-1)/n = 5
(1-1/n) = 5/3
1/n = 1-5/3
1/n = -2/3

n = -3/2 ......................

From x = 3/4n = 3/4(1/n) = 3/4(-2/3)

x = -1/2 ......................

Hence the series is (1 + x)^n = (1 - 1/2)^(-3/2) = 1 + 3/4 + 3•5/4•8 + 3•5•7/4•8•12 + ...
Solving (1 - 1/2)^(-3/2) gives 2(2)^(1/2) meaning 2 * square root of 2

Thanks a lot bro

yea, i made the error while typin, u postd ur comment while i was editin mine, its already correctd.
nice one!
So u 're in possession of d solution afterall

Laplacian:
yea, i made the error while typin, u postd ur comment while i was editin mine, its already correctd.
nice one!
So u 're in possession of d solution afterall

D one I solve is different from d one u solved, and its a bit easier.
Thanks man

Pat £inst£in:

D one I solve is different from d one u solved, and its a bit easier.
Thanks man

k, we know that, for a series to be representable in a binomial form, the terms must have a recurrent structure, and the numerator & denominator must have the same number of factors. The two conditions are satisfied for case u solved, since
T_n=#(2r+1)/[4^n-1(n-1)!]=T_n-1*(2n+1)/[4^n-1(n-1)!]
the one i solved also to satisfies:
T_n=#(2r+3)/[3ⁿ⁺¹*(n+1)!]=T_n-1(2n+3)/[3ⁿ⁺¹*(n+1)!], but it fails for the nunber of factors, to accommodate that, consider the eqn;
(1+x)ⁿ=1+nx+[n(n-1)/2!]x²+[n(n-1)(n-2)/3!]x³+...
I think a little modification can make it suit the conditions
e.g
(1+x)ⁿ-(1+nx)=[n(n-1)/2!]x²+[n(n-1)(n-2)/3!]x³+...
We divide thru by nx to obtain
[(1+x)ⁿ-(1+nx)]/nx=[(n-1)/2!]x+[(n-1)(n-2)/3!]x²+[(n-1)(n-2)(n-3)/4!]x³+...
Use ur method for the last eqn lets c hw far it goes...

hmmm..what more can i say..?
my able gurus have done justics...

pls kindly try the questions above..

futurewise11: a)a person wants to make annual deposits for 25 years into an ordinary annuity that earns 9.4% compounded annually in order to then make 20 equal annual withdrawals of $25,000, reducing the balance in the account to zero. How much must be deposited annually to accumulate sufficient fund to provide for this payments? How much total interest is earned during this entire 45 year process?


b) A business borrows $80,000 at 15% interest compounded monthly for 8 years. What is the monthly payment? What is the unpaid balance at the end of the first year? How much interest was paid during

these are the answer,
a) A = $2467.96; interest $438,301;
b) monthly payment $1435.63; unpaid balance at the end of the first year $74397.48;but we are to show the workings


have actually try to solve it but am arriving at something different. for ai)i got 5,562.11 using annuity/sinking fund formulae.

please great maths guru of nairaland kindly help in resolving it.

please help to resolve this misery

Bro, be at peace, your questions are quite easy to solve, just that they're too long. I'm feeling reluctant but I'll try post the solutions tonight.
------------------
Cheers,
Ortarico.

Laplacian:
k, we know that, for a series to be representable in a binomial form, the terms must have a recurrent structure, and the numerator & denominator must have the same number of factors. The two conditions are satisfied for case u solved, since
T_n=#(2r+1)/[4^n-1(n-1)!]=T_n-1*(2n+1)/[4^n-1(n-1)!]
the one i solved also to satisfies:
T_n=#(2r+3)/[3ⁿ⁺¹*(n+1)!]=T_n-1(2n+3)/[3ⁿ⁺¹*(n+1)!], but it fails for the nunber of factors, to accommodate that, consider the eqn;
(1+x)ⁿ=1+nx+[n(n-1)/2!]x²+[n(n-1)(n-2)/3!]x³+...
I think a little modification can make it suit the conditions
e.g
(1+x)ⁿ-(1+nx)=[n(n-1)/2!]x²+[n(n-1)(n-2)/3!]x³+...
We divide thru by nx to obtain
[(1+x)ⁿ-(1+nx)]/nx=[(n-1)/2!]x+[(n-1)(n-2)/3!]x²+[(n-1)(n-2)(n-3)/4!]x³+...
Use ur method for the last eqn lets c hw far it goes...

I tried it but couldn't solve further to arrive at an answer; let's leave the answer at: Tn=[(3n+2)!/3!]/(2n/2•[(3n+2)/2]!)/[3n+1•(n+1)!]
Where u stopped initially

Thanks a lot, I really appreciate

benbuks: ^ ok...m on bed nw..mayb i ‘d try dem 2maro...quit cheep sha.

I'm still waiting for ur solutions Sir

Prove that tan^-1(x+z) = tan^-1(x) + 2sin(z)sin(z) - [(2sin(z))^2sin^2(z)]/2 + [(2sin(z)^3sin^3(z)]/3 - ...
Where cot(z) = x

It now remains the solution for this last question, but u can still post ur solutions 4 d oda questions if they are different from d ones posted

Thanks

Futurewise11,
I'm sorry, I've to post the solution to the B part this night, when the day breaks, I promise to post that of the A. Here we go:

B(i). To solve, we shall use the formula for amortization. This is because it deals with a borrowed amount/loan.

P = V[1 - (1 + i)^-n/i]^-1
The monthly payment, P is needed to pay-off the debt,
Where; V = $80,000
i = 0.15/12 = 0.0125
n = 8*12 = 96

Applying the formula above:
P = 80,000[1 - (1 + 0.0125)^-96/0.0125]^-1
P = 80,000[1 - (1.0125)^-96/0.0125]^-1
P = 80,000[1 - 1/(1.0125)⁹⁶/0.0125]^-1
P = 80,000[1 - 0.3034428/0.0125]^-1
P = 80,000[0.6965571/0.0125]^-1
P = 80,000*0.0125/0.695571
P = 1,000/0.695571
:. ~ P = $1,435.63

B(ii). Unpaid balance after the first year (UB): V(1 + i)ⁿ - P[(1 + i)ⁿ - 1/i]

Where; V = $80,000
i = 0.0125
P = $1,435.63
n = 12


When you substitute those values you get:
UB = 92860.3614 - 18462.7207
:. ~ UB = $74,397.64
-------------------
Cheers,
Ortarico.

here i am clinic's. some 1 should pls explain mathematical induction 2 me with examples. thanks in anticipation

Pat £inst£in:
Sum the series:
[5/(3 • 6)] + [(5 • 7)/(3 • 6 • 9)] + [(5 • 7 • 9)/(3 • 6 • 9 • 12)] +...~

NB: the • means multiplication as in, 2 • 3 = 6

Thanks @aysuccess99

The answer is -2/3 + sqrt 3
The process just too long

akpos4uall:

The answer is -2/3 + sqrt 3
The process just too long

Pls sir, try and post the solution for us

pls help me with this question..av bin tryin since....
lim. [(x^4 - a^4)/x-a]
x-->0

Pat £inst£in:
Sum the series:
[5/(3 • 6)] + [(5 • 7)/(3 • 6 • 9)] + [(5 • 7 • 9)/(3 • 6 • 9 • 12)] +...~

NB: the • means multiplication as in, 2 • 3 = 6

Thanks @aysuccess99

Let me give it a try
(1 + x)ⁿ = 1 + nx + n(n - 1)x²/2! + n(n - 1)(n - 2)x³/3! + n(n - 1)(n - 2)(n - 3)x⁴/4!
+ ...
This can be written as
(1 + x)ⁿ - 1 - nx = n(n - 1)x²/2! + n(n - 1)(n - 2)x³/3! + n(n - 1)(n - 2)(n - 3)x⁴/4! + ... Eqn (1)

Let z = 5/(3 * 6) + 5 * 7/(3 * 6 * 9) + 5 * 7 * 9/(3 * 6 * 9 * 12) + ...
Multiply both sides by 3

3z = 3 * 5/(3 * 6) + 3 * 5 * 7/(3 * 6 * 9) + 3 * 5 * 7 * 9/(3 * 6 * 9 * 12) + ...
3z = 3 * 5 * (1/3)²/(1 * 2) + 3 * 5 * 7 * (1/3)³/(1 * 2 * 3) + 3 * 5 * 7 * 9 * (1/3)⁴/(1 * 2 * 3 * 4) + ...

3z = 3 * 5 * (1/3)²/2! + 3 * 5 * 7 * (1/3)³/3! + 3 * 5 * 7 * 9 * (1/3)⁴/4! + ... Eqn (2)

Comparing eqn (1) & eqn (2)
3z = (1 + x)ⁿ - 1 - nx Eqn (3)

3 * 5 * (1/3)² = n(n - 1)x²
5 = 3n(n - 1)x² Eqn (4)

3 * 5 * 7 * (1/3)³ = n(n - 1)(n - 2)x³
35 = 9n(n - 1)(n - 2)x³ Eqn (5)

3 * 5 * 7 * 9 * (1/3)⁴ = n(n - 1)(n - 2)(n - 3)x⁴
35 = 3n(n - 1)(n - 2)(n - 3)x⁴ Eqn (6)

Divide equation (5) by equation (4)
7 = 3x(n - 2) Eqn (7)

Divide equation (6) by equation (5)
1 = x(n - 3)/3
9 =3x(n - 3) Eqn (

Divide equation ( by equation (7)
9/7 = (n - 3)/(n -2)
9(n - 2) = 7(n - 3)
9n - 18 = 7n - 21
2n = -3
n = -3/2
Substitute -1.5 for n in equation (
9 = 3x(-1.5 - 3)
9 = 3x * -4.5
x = 9/(3 * -4.5)
x = -2/3 , n = -3/2

Remember z is what we are looking for. Hence substitute the obtained values of n & x into equation (3)
3z = (1 + x)ⁿ - 1 - nx Eqn (3)
3z = (1 + -2/3)^(-3/2) - 1 - (-2/3)(-3/2)
3z = (1/3)^(-3/2) - 1 - 1
3z = 3^(3/2) - 2
3z = (3^3)^(1/2) - 2
3z = 27^(1/2) - 2
z = 3 * 3^(1/2)/3 - 2/3
z = 3^(1/2) - 2/3

akpos4uall:

Let me give it a try
(1 + x)ⁿ = 1 + nx + n(n - 1)x²/2! + n(n - 1)(n - 2)x³/3! + n(n - 1)(n - 2)(n - 3)x⁴
+ ...
This can be written as
(1 + x)ⁿ - 1 - nx = n(n - 1)x²/2! + n(n - 1)(n - 2)x³/3! + n(n - 1)(n - 2)(n - 3)x⁴ + ... Eqn (1)

Let z = 5/(3 * 6) + 5 * 7/(3 * 6 * 9) + 5 * 7 * 9/(3 * 6 * 9 * 12) + ...
Multiply both sides by 3

3z = 3 * 5/(3 * 6) + 3 * 5 * 7/(3 * 6 * 9) + 3 * 5 * 7 * 9/(3 * 6 * 9 * 12) + ...
3z = 3 * 5 * (1/3)²/(1 * 2) + 3 * 5 * 7 * (1/3)³/(1 * 2 * 3) + 3 * 5 * 7 * 9 * (1/3)⁴/(1 * 2 * 3 * 4) + ...

3z = 3 * 5 * (1/3)²/2! + 3 * 5 * 7 * (1/3)³/3! + 3 * 5 * 7 * 9 * (1/3)⁴/4! + ... Eqn (2)

Comparing eqn (1) & eqn (2)
3z = (1 + x)ⁿ - 1 - nx Eqn (3)

3 * 5 * (1/3)² = n(n - 1)x²
5 = 3n(n - 1)x² Eqn (4)

3 * 5 * 7 * (1/3)³ = n(n - 1)(n - 2)x³
35 = 9n(n - 1)(n - 2)x³ Eqn (5)

3 * 5 * 7 * 9 * (1/3)⁴ = n(n - 1)(n - 2)(n - 3)x⁴
35 = 3n(n - 1)(n - 2)(n - 3)x⁴ Eqn (6)

Divide equation (5) by equation (4)
7 = 3x(n - 2) Eqn (7)

Divide equation (6) by equation (5)
1 = x(n - 3)/3
9 =3x(n - 3) Eqn (

Divide equation ( by equation (7)
9/7 = (n - 3)/(n -2)
9(n - 2) = 7(n - 3)
9n - 18 = 7n - 21
2n = -3
n = -3/2
Substitute -1.5 for n in equation (
9 = 3x(-1.5 - 3)
9 = 3x * -4.5
x = 9/(3 * -4.5)
x = -2/3 , n = -3/2

Remember z is what we are looking for. Hence substitute the obtained values of n & x into equation (3)
3z = (1 + x)ⁿ - 1 - nx Eqn (3)
3z = (1 + -2/3)^(-3/2) - 1 - (-2/3)(-3/2)
3z = (1/3)^(-3/2) - 1 - 1
3z = 3^(3/2) - 2
3z = (3^3)^(1/2) - 2
3z = 27^(1/2) - 2
z = 3 * 3^(1/2)/3 - 2/3
z = 3^(1/2) - 2/3

Cool approach there

Thanks a lot

Pat £inst£in:

Cool approach there

Thanks a lot

...pls try my questions sir....

Pat £inst£in:

Cool approach there

Thanks a lot

You are welcome.
My pleayure...

benbuks: pls help me with this question..av bin tryin since....
lim. [(x^4 - a^4)/x-a]
x-->0

Check well I feel it should be lim as x-->a not 0

(x⁴ - a⁴)/(x - a)
(x² - a²)(x² + a²)/(x - a)
(x - a)(x + a)(x² + a²)/(x -a)
(x + a)(x² + a²)
To get the limit as x-->a, fix x as a into the above expression
(a + a)(a² + a²)
2a*2a²
4a³
If it were to be limit as x-->0 as asked, fix x as 0
(0 + a)(0² + a²)
a * a²
a³

@akpos4uall, nice work. Happy New Year bruv!

doubleDx: @akpos4uall, nice work. Happy New Year bruv!

Thank you & you too.

akpos4uall:
You are welcome.
My pleayure...


Check well I feel it should be lim as x-->a not 0

(x⁴ - a⁴)/(x - a)
(x² - a²)(x² + a²)/(x - a)
(x - a)(x + a)(x² + a²)/(x -a)
(x + a)(x² + a²)
To get the limit as x-->a, fix x as a into the above expression
(a + a)(a² + a²)
2a*2a²
4a³
If it were to be limit as x-->0 as asked, fix x as 0
(0 + a)(0² + a²)
a * a²
a³

.

on point sir..hw did u kw its ‘a‘ guess u av d txt buk......bt it was written zero....thanks alot...

kindly try d former questions pls...