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

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

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

smurfy: Happy New Year To Y'All!

So grateful to God for a successful crossover.

Welcome, welcome, welcome to 2014!!!

benbuks: ....welcome bro ..hope u came with my new year gifts....

main while, dia sm unsolved enigmas in d previous pages....help us out if u could pls....mine inclusive....tnx broo..

1 love

smurfy:

Happy new year, benbuks!

About your gift... The big chicken I reserved for you dis-integrate-d during cooking on Jan 1. I'm so, so sorry.

smurfy:

Happy new year, benbuks!

About your gift... The big chicken I reserved for you dis-integrate-d during cooking on Jan 1. I'm so, so sorry.

jaryeh: Happy New Year to you all my masters!

smurfy: Where's Richiez? I thought he'd be the first to welcome us all into the new year.

**fuming**

Maybe he's yet to return from his village trip. Granny and grandpa won't let him...

benbuks: @smurfy r u bk to sch?

akpos4uall:

Here is an example for you www.nairaland.com/1147658/nairaland-mathematics-clinic#13657469

benbuks:

aka m.i

benbuks: ...welcome back..our outstandin champion.......ur frequency on. the thread suddenly became a null set...wetin hapun na?
abi ur “wife“ born ni..?

jaryeh:

ha! Oga mi. Good evening sir.
Na work things jare.... I promise to always show up whenever I'm less busy.

benbuks: 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


.

akpos4uall: ^^^
Instead of those plenty plenty wahalai above for problem 3, we can just state that the (r + 1)th term in the expression (1 + x)ⁿ is ⁿC_r * x^r
Where r = 0, 1, 2, ... n
ⁿC_r = n!/(r! * (n - r)!)
:. The (r + 1)th term of x²ⁿ(1 + x^-2)²ⁿ = ²ⁿC_r * x^-2r * x²ⁿ
= ²ⁿC_r * x^{2n - 2r}
From here we can arrive at the right answer as posted earlier.

akpos4uall: ^^^
Instead of those plenty plenty wahalai above for problem 3, we can just state that the (r + 1)th term in the expression (1 + x)ⁿ is ⁿC_r * x^r
Where r = 0, 1, 2, ... n
ⁿC_r = n!/(r! * (n - r)!)
:. The (r + 1)th term of x²ⁿ(1 + x^-2)²ⁿ = ²ⁿC_r * x^-2r * x²ⁿ
= ²ⁿC_r * x^{2n - 2r}
From here we can arrive at the right answer as posted earlier.

akpos4uall:
Problem 1
n(n + 1)(n + 2)...(n + k)
1 * 2 * 3 * ...(n - 1)(n)(n + 1)(n + 2)...(n + k)/(1 * 2 * 3 * ...(n - 1)
(n + k)!/(n - 1)!


Problem 2
1² * 2² * 3²...n²
(1 * 1 * 2 * 2 * 3 * 3...n * n)
(1 * 2 * 3...n)(1 * 2 * 3...n)
n! * n!
(n!)²

Problem 3
find the term independent of x in. [x + 1/x]^2n
(x + 1/x)^2n = (x(1 + x^-2))²ⁿ
=x²ⁿ(1 + x^-2)²ⁿ
From the binomial expansion,
(1 + x)ⁿ = 1 + nx + ⁿC₂ * x² + ⁿC₃ * x³ + ⁿC₄ * x⁴... + ⁿC_n * xⁿ
Where ⁿC_r = n!/(r! * (n - r)!)
:. x²ⁿ(1 + x^-2)²ⁿ = x²ⁿ[1 + 2n(x^-2) + ²ⁿC₂ * (x^-2)² + ²ⁿC₃ * (x^-2)³ + ²ⁿC₄ * (x^-2)⁴... + ²ⁿC_2n * (x^-2)²ⁿ]
= x²ⁿ[1 + 2n(x^-2) + ²ⁿC₂ * x^-4 + ²ⁿC₃ * x^-6 + ²ⁿC₄ * x^-8... + ²ⁿC_2n * x^-4n]
= x²ⁿ + 2n(x^{2n -2}) + ²ⁿC₂ * x^{2n -4} + ²ⁿC₃ * x^{2n -6} + ²ⁿC₄ * x^{2n -8} + ... + ²ⁿC_2n * x^{2n -4n}
The (r + 1)th term of the above expression can be written as
²ⁿC_r * x^{2n -2r} for example the 4th term will be gotten when r = 3 which is ²ⁿC₃ * x^{2n -6}
Note: 1 = ²ⁿC₀
2n = ²ⁿC₁
In order the get the term independent of x, the power of x from the expression = 0
==> 2n - 2r = 0
2n = 2r
n = r
The coefficient =
²ⁿC_r =
²ⁿC_n
The term independent of x in [x + 1/x]^2n
= (2n)!/((n!)(2n - n)!)
= (2n)!/(n! * n!)

Pat £inst£in:

Pls help me with this

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

benbuks:

sorry i could‘nt. solve ur question.....

did‘nt av tym to look into it well....

mayb i dont understand it well..
al d same ..our able gurus will do justice....tnx 4ur understandin..

1love

Pat £inst£in:

Pls help me with this

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
Or clearer,
Let 2sin(z) = y
We now have
Prove that tan^-1(x+z) = tan^-1(x) + ysin(z) - (y^2/2)sin^2(z) + (y^3/3)sin^3(z) - (y^4/4)sin^4(z) + ...
Where cot(z) = x
(Question under TAYLOR'S EXPANSION)

Thanks

Laplacian:
i thought u 've resolved it:
now,
f(x,z)=f(x)+zf'(x)+z²f"(x)/2!+...

If f(x,z)=tan^-1(x+z), then the above equation resolves into
tan^-1(x+z)=tan^-1x+zf'(x)+z²f"(x)/2!+...
Since f(x)=tan^-1x
then,
f'(x)=1/(1+x²)

f"(x)=-2x/(1+x²)²=-2x[f('x)]²
f'"(x)=d(-2x[f('x)]²)/dx, e.t.c
now since cot(z)=x, => cot²(z)=x², or 1+x²=1+cot²z=cosec²x,
:. sin²x=1/(1+x²),
i hop u can complet the rest

*i must comment that, your series is slightly INCORRECT*

rashywire: pls help with these questions

1. if x and y are real, solve d equation;
jx / 1+jy = 3x+j4/ x+3y
SOLUTION::
first cross multiply u wud get:
jx^2+j3xy=3x+j4+j3xy+j^24y....
=>>jx^2=3x+j4-4y..
=>>jx^2-3x=j4-4y.
by comparison x^2=4 & 4y=3x.
x=+/-2 and y=+/-3/2

rashywire:
2. if z = a+jb / c+jd , where a,b,c and d are real quantities, show d@
(a) if z is real then a/b = c/d and

(b) if z is entirely imaginary then a/b = -d/c