₦airaland Forum

Swagalord18: ur very good ooh ...kudos
nd i lyk d way u take ur tym 2 make use of the sub nd sup 4 beta understanding.
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[size=16pt]Take this as desert[/size]
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The nth term of the series 5 + 5 + 6¹/₂ + ..... Is given by T_n = a(¹/₃)^n-1 + bn. Find the values of a and b, and the sum of the first n terms of the series

Swagalord18: corrrect ma'am
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[size=16pt]Here's lunch[/size]
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if T_n+1 = a + bT_n for a sequence T₁, T₂, T₃, . . ., show that T_n+1 - T_n = b^n-1 [a - (1 - b)T₁]

SMH!!!

It's obvious you are not in need of the solutions. I'm equally not here to impress you or anybody.

Take back your meals and feed yourself first.

At least, I know you are very skinny, and in need of sumptuous meals.

I'm out!!!

wisemania: mth125....fym u be super star...kip it up,let me go n sip some shekpe and some dogo b4 i strt my own display.....

hey bro, no blame me much for solving that small question sha.

I am acting on instructions by my 5-STAR GENERALS sha. I solve the moderate ones. I refer only the tougher questions to them. They are the best.

Swagalord18: ur very good ooh ...kudos
nd i lyk d way u take ur tym 2 make use of the sub nd sup 4 beta understanding.
.
Take this as desert
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The nth term of the series 5 + 5 + 6¹/₂ + ..... Is given by T_n = a(¹/₃)^n-1 + bn. Find the values of a and b, and the sum of the first n terms of the series

Hey bro, check this, from where you copied it out. It should be 6¹/₃.

Having said that, put n=1 and n=2 in the T_n equation and equate to 5 and 5 respectively.

You will then get the two equations

a+b=5 . . . . . . . . . . . . . . . . . . . (1)
^a/₃+2b=5. . . . . . . . . . . . . . . . .(2)

solve these simultaneously to get
a=3, b=2

substitute these in the T_k equation(I am calling it T_k instead of T_n because I want to sum from k=1 to n. Remember they are dummy variables after all )

T_k= 3(¹/₃)^k-1 + 2k

sum over all n and simplify (you should use the sum of a GP formula on the first summand and the sum of an AP formula on the second summand here).

The tiny details are left as an easy exercise for the reader

Final answer is

S_n = ⁹/₂ [1-(¹/₃)ⁿ ] + n(n+1)


Thats all!

benbuks: ..do, than say.

integrate first, you get
y'=gt+c₁
y' is velocity and at t=0, the initial velocity y'(0) is 0 since the particle is falling from rest. Substitute these facts to get c₁=0.
Thus,
y'=gt.
Integrate again to get
y=gt^2+c₂

but at t=0, the distance covered is zero. Substitute these into the above equation to get c₂=0.

Thus,

y= gt^2/2

thats all!

2nioshine: ....nice one...2 dat ur quest abt the set... she is currently in her finals..maths dpartmnt ...guess itz nt u ryt?

nah.

2nioshine: here is it...
Take T(n)=T subscript nd S(n),S(n-1)=s sub n&S sub (n-1) respectively
generally T(n)=S(n)-S(n-1)...an establishd fact
also if S(n)=n^2
:- S(n-1)=(n-1)^2
hence d above gvs
T(n)=[n^2]-[n-1]^2
when n=1(u can chk any chosen no)
substn we have
T(1)=1-0=1
T(5)=25-16=9...hope it helps

smarter solution.

Guess not all will understand the step sha, especially the fact that Tn=Sn-(Sn-1).

With this method, the formula for the nth term is gotten "free"

benbuks: If we drop a stone,we can assume air resistance ("drag" to be negligible. Experiments show that under that assumption the acceleration y"= d^2 y/dt^2 of this motion is constant (equal to the so-called acceleration of gravity g=9.80 ms^-2 =32ftsec^-2) . State this as an ODE for y(t), the distance fallen as a function of time. Solve the ODE to get the familiar law of free fall, y=gt^2 / 2

na to just dey integrate, substitute, and apply initial conditions.

It is not challenging enough

Hi doubleDx

what's the name and author of that textbook you uploaded while solving that paraboloidal stuff?

Swagalord18: prove that if the first term of an A.P is 1, and the sum of n terms is n², the A.P is 1,3,5.....9

the formula for the sum of n terms of an AP is

Sn=ⁿ/₂[2a+(n-1)d]
where a is the first term and d is the common difference.
Therefore by putting a=1(given), we have
Sn=ⁿ/₂[2+(n-1)d]

Equate this sum to n²(given) and simplify(since n not equal to 0), to get
n= ¹/₂[2+(n-1)d]
therefore
2n=2+(n-1)d
2n=2+dn-d
2n-dn=2-d
n(2-d)=2-d
n(2-d)-(2-d)=0
(n-1)(d-2)=0
from which we have
d-2=0
d=2

first term a=1 (given)
2nd term =a+d=1+2=3
3rd term=a+2d=1+2(2)=5
and so on.

Thats all!

smurfy: Solve the simultaneous equations
(10^x)(4^y) = 1
8^x = 10^(y+1)

taking Log₁₀ of both sides, we have

x+yLog4=0. . . . . . . . . . . .1
xLog8-y=1. . . . .. . . . . . . .2
from eqn 1, x= -yLog4. Substitute this for x in Eqn 2
-(Log4)(Log8)y-y=1

thus,
y= -¹/_{(1+ Log8 Log4)}

x=-yLog4= ^Log4/_{(1 + Log8 Log 4)}



thats all!

Mr Calculus: hahaha......na because of fail way i fail am 9 make u dy writ all those vocalbs?....
i dint espect to b right,dats y i posted it.....

the question I answered is even 3 times tougher than what you asked.

Your question cheap sef.

1/(cuberoot2)-1= [ (2^2/3)/₂ ] - 1

Mr Calculus: hmm.......@jackpot..
first of all u change my ques from -16 to -16+oi....i followed that.
then later to 16e^i(pi) and w^4...... i was totally loss by dat move cause u ignored d -ve sign attach to d 16 .....
ALL D SAME DIS IS D ANSWER,BUT WAT I WANT IS D SOLUTION::
root2(1+j),,root2(-1+j),,root2(-1-j),,root2(1-j).....
PLZ WEN SOLVIN TRY TO USE SIMPLE TERMS CAUS ME NO TOO GUD 4 MATHS CAUS AM STILL A FRESHER.....
TNX IN ADVANCE@ALL MY GENERALS PLZ HELP ME OUT WIT D SOLUTION.

guy, e^i(pi)=-1.

Sorry, your high-end username misled me into thinking that you can be able to fill in the gaps

well, let me be a little bit explanatory.

Let T=theta

then the Euler formula is e^iT=cos T+ i sin T

its easy to see that e^i(pi)=cos pi + i sin pi=-1+i0=-1

and so, -16=16(-1)=16e^i(pi). Hope you now get it?

then, about that w^4,

if w is a fourth root of a complex number z (i.e., w=z^1/4), then, w^4=z. Simple elementary algebra.

So, if you use the Euler number on my first answer, you will have
2e^i(pi/4)= 2(cos pi/4 + i sin pi/4)
=2[(root2)/2+i(root2)/2 ]=root2 + iroot2
= root2 (1+i)

did you notice that it correspond with your first answer?

I'll leave the remaining three for you to simplify in rectangular form as I just did. They should equally correspond with what you have there.

Sorry, I'll apologize once again that your username misled me into assuming that you are good in Math. My bad!!!

Mr Calculus: & DAT MAKES UR ANSWER WRONG......

it doesnt make my answer wrong, since I didn't answer your question

hope you got my drift.

WEN I DID IT I GOT
-1/(2^1/3+2^2/3+2)
but i dont know if am right.
plz try again and others plz help out

you are totally, entirely, capitally and wholistically wrong since there are serious radicals in your denominator.

You started with a question having only one radical, and your wrong answer had 2 radicals in the denominator!!!

labodinho: Jackpot scares me off here i must say...


A lady??

hmmm!

shhhhh. . . Easy oooo.

The questions I solve are often very cheap left-overs. You should be giving credit to my fellow hommies and 5-STAR Generals who left them for me to use and shine.

They solve the harder ones. More credits to them

I-am-Winner:
there are certain tins in maths that I cant forget.. Smtin lyk dis..

Ans is (root 2) + 1

hehehe. Bros, no be square root oooooo Na cube root.

they hardly teach this type of rationalization for high school sef.

Bro, i like your username oooo

Mr Calculus: rationalize:
1/(cuberoot2) -1

i hope you meant ¹/_{[(cuberoot2)-1]}

Now, since
¹/_(x-1)=(x²+x+1)/_(x³-1),

we have (upon setting x=cuberoot2=2^1/3) that

¹/_{[(cuberoot2)-1]}

equals

(2^2/3+2^1/3+1)/_(2^3/3-1)

which simply equals

2^2/3+2^1/3+1.

Laplacian: ...my mistake...

Mr Calculus: PLZ SIRS NO BODY HAS PROVIDED A SOLUTION TO DIS.
U GUYS SHOULD PLZ HELP ME OUT

in the form a+ib
z=-16+0i
[s]sorry, Mathematicians dont write the imaginary part with j like Physicists and Engineers [/s]

now, for the fourth root, i'll prefer working with the exponential form to avoid the impossible task of typing out square roots here.

z=-16+0i=16e^i(pi)=w⁴

Using the nth root of a complex number formula

w=nth root of |z| times e^i[(Argz+2k pi)/n] for k= 0, 1, 2 ,. . ., n-2, n-1

Thus, the fourth roots are simply

w= 2e^i(pi/4), 2e^i(3pi/4), 2e^i(5pi/4), 2e^i(7pi/4)

Alpha Maximus: I don't think a negative number such as -16 has a root

bro, you still far oooooo. It has ooooo. Na elementary complex number stuff. MAT101 in my school.

Laplacian: ...@Maximus, d rational mathematician...till dusk...u'll c d solution...

my friend, the rational Alpha Maximus no even decode the sarcasm sef

rhydex 247: @ jackpot i tried to modify my error yesterday nite but d network hook me. Thanks 4 d correctn.
. . .

you're welcome, always

2nioshine: @Jackpot u seems like my one tym gal who actually gave me a tough time and made me study like c.obi (hope u r nt d unizik gal) ...alpha i luv ur math spirit no doubt u r no baby mathematician so lets b a little r***tional here(seun no dy pay or give us grades abeg),...double dx i was kinda di***pointed,but now u apologised we r kul...u remainent my most respectd boss...# Still in my observation mood#....nice wuk @ all 4 ur tym nd contributn(wish i could list al ur names)..sti following...wait o! Where is Richez

which set?

bro rhydex, this is where you defined your multiplication.

rhydex 247: here is the question again
Show that V=IR^2 is not a vector space over IR with respect to operation (a,b)+(c,d)=(a+c, b+d) and k(a,b)=(ka,kb)

why did you use k(a,b)=(ka,b) in your solution?

SMH

rhydex 247: soln.
Suppose u=(3,4). If r & s € k. r=1 and s=2.
(r+s)u=ru+su.
(r+s)u=(1+2)(3,4)=(9,4).
ru+su=1(3,4)+2(3,4)
ru+su=(3,4)+(6,4)= (9, . Since (r+s)u is not equal to ru+su. Hence is not a vector space.

in your question, you defined multiplication

k (a,b)=(ka,kb)

thus, su= 2(3,4)=(2*3,2*4)=(6,8 )

why did you erroneously write your su as (6,4)?

Also, (1+2)(3,4)=3(3,4)=(3*3,3*4)=(9,12)

why did you also write the result as (9,4)?

I taya oooo

I repeat IR² is a vector space with the definition of multiplication you gave!!!

Laplacian: ...@Jackpot...i'll advic u 2 reserv all comments concernin dat gentleman, i belive by now it should b obvious 2 everyone dat he abhors criticism, more developd in rainin insults dan coherently providin answers 2 questions...Jackpot i want 2 ask u a question, here's a man who claims 2 promot knowledge...publicly postin insults, what kind of knowledg is he promotin?...check his record from d tim he had issue wit Doubledx, about 75% of his issues seems 2 lie in d title dat some ppl just chose 4 demselves (just 4 fun)...i sugest we let him chose a name 4 himself 2 stop all dis cat & rat...my concern is mostly 4 our youngstars...by d way wat's 'SMH'?...

I hear you, bro!





SMH means "Shakes My Head"

^you cajoled the other person saying his solution is trial and error and that you are going to do it direct. Here you are saying that the Remainder Theorem you applied is trial and error method.

Okay ooo

Now, why didnt you get y=2 when applying your theorem? I suspect super-film trick here

If you get y=3, why not substitute into the more simpler x+y=5 to get your value of x?

I repeat, the remainder theorem is not about trial and error. You may divide a polynomial with x-2 and get a remainder. You may choose to divide the same polynomial with x-5 and get another remainder.

Maybe you wanted to say Factor Theorem.

But still before you apply both, f(x) must be a polynomial and what you used is not a polynomial.

The sky is your limit though, only if you will accept your mistakes, be less-abusive and learn. You can still be the future Udeme guiness man of Mathematics

rhydex 247: lol @ jackpot u said d questn is nt correct. I dnt expect dat 4rm u. Anyway. All is well.
THE BEST DNT ALWAYS DO THE BEST.

i will give a definition of a vector space.

Let X be a non-empty set, K- a scalar field(usually K is taken to be IR, the real line). Suppose the functions +, . ,
+ : X x X ----> X
. : K x X ----> X
called respectively addition and scalar multiplication are defined(i.e., for arbitrary x,y € X and scalar k€K, x+y€X and k.x€X such that
1. X is an abelian(or commutative) group, i.e., for arbitrary x,y,z€X, x+y=y+x; x+(y+z)=(x+y)+z, there exists 0€X such that x+0=x. For each x€X, there exists (-x)€X such that x+(-x)=0.

2. k.(x+y)=k.x+k.y for each x,y€X, k€K.

3. (a+b).x=a.x+b.x for all a,b€K, x€X

4. (ab).x=a(b.x) for each a,b€K, x€X.

5. 1.x=x for 1€K, x€X

then X is called a vector space(some author calls it linear space) over K. Moreover, if K is the set of real numbers, then X is called a real vector space or a real linear space.


Functional Analysis stuffs.

Laplacian: ...@Jackpot...fear no allow me talk ooo...

I didnt know you noticed his miscellaneous errors too oh. And you are waiting for me to speak out, bah? Dude did magic and 3 appeared. He even substituted y=3 only in the indices/powers, leaving the y below untouched. Wonderful.

Now, I see what you mean. Dude is too insultive and yet, so error-prone!

Dude needs to be demoted from the rank of a Mathematics Sergeant to a Mathematics Corporal and with half-pay salary for all his mathematical atrocious atrocities.

last statement na joke ooo, before him go scatter my head with verbal missiles

Alpha Maximus: ..... Smh, fucktards everywhere and a mega person(jackpot) here!! Can you now see why there's always an issue of 'half-baked' grraduates? I'm sure you never even made it to the oven!!! You're asking me if x^x+y^x=31 is a polynomial!! You're are an illiterate of no comparison!!*salutes* Wasn't it (5-y)^2+y^2-31=0 that I used as a polynomial!! For your info, a polynomial is an expression constructed of variables and constants using operations of addition, subtraction and multiplication with non-negative integer exponents(culled from Wikipedia)....do you see any mention of a polynomial being stictly of the 3rd or 4th degree, you intellectually devoid slowpoke!!! And about why I used the remainder theorem, isn't the first step of using it trying out integer values of 'x' or'y' first? Calling me a learner, you must be truly insane!! You think complex looking questions are always the hardest to solve, irredeemable mugu!! Who is forming,online again? Weren't you the one saying your solution was sexy? Pride is one of the deadly sins oh, and your seat is being reserved in the VIP section of Hell!!! And about that 'turns my solution into tissue paper' comment, the last time I met something like you , I flushed it down the toilet!!! Pompous piece of shit

you applied the remainder theorem on precisely the equation (5-y)^(5-y)+y^y=31.

A polynomial has only integral powers. The equation you applied your newfound cu.m-fake "Alpha Maximum-Remainder Theorem" is not a polynomial. The general form of a polynomial is
P(x)=a_n x^n+. . .+a_2 x^2+a_1 x +a_0.

Bros, remainder theorem says that if a polynomial f(y) is divided by y-a, the remainder is R=f(a). For Gods sake, the remainder theorem is specifically talking about the remainder!

Dude even checked Wiki for definition of a polynomial which he carelessly misinterprets.

For the last time (5-y)^(5-y)+y^y=31 is not a polynomial.

Alpha Maximus: ....No offence, but I have to say the answer was kind of obvious! Though the 'trial and error' method worked, it becomes impractical when a similar question has much higher values of 'x' and 'y'.....thus, it would be better to stick to strictly analytical , non-graphical and algebraic processes AKA the HARD NONSENSE and more directquasi-direct way....
This is my own solution:
X+y=5, x^x+y^X=13
From(I), x=5-y(3), input (3) into (2)
(5-y)^(5-y) +y^(5-y)=13
Now converting the rather tedious equation into function form, we have:
F(y)=(5-y)^(5-y) + y^(5-y)-13=0
Using the remainder theorem, we have that y=3
Since the use of the long division method would still leave the index/power of (5-y) unknown, I will substitute y=3 strictly into the index and convert the equation back to algebraic form giving:
(5-y)^(5-3)+y^(5-3)-13=0
(5-y)^2+y^2-13=0
Expanding and collecting like terms, we have a quadratic expession of:
2y^2-10y+12=0
Factorising we have:
(2y-4)(y-3)=0
Thus,y=3(already established) and,
2y=4, y=4/2=2(this is also the value of 'x')
Therefore, x=2, y=3
Though this is more tedious than your method, it is more practical when larger values are involved

The bolded is just cock-and-bull.

First of all, what does the remainder theorem say?

Once a polynomial f(x) is divided by x-a, the remainder is another polynomial of 1 degree less than that of f(x) together with a remainder R still to be divided by x-a.

i.e., f(x)/(x-a)=g(x)+R/(x-a)
therefore f(x)=(x-a)g(x)+R
substituting x=a yields that
the remainder R=f(a)

back to the question. What is a polynomial? I would leave that for you to find out.

The equation x^x+y^y=13 is not a polynomial.

Why then are you applying the remainder theorem which specifically talks about polynomials?

And the funny thing is that you said the application yields y=3. SMH

Rubbish. Bros, no just dey form anyhow, you are still a BIG learner.

Converts his answer sheet to toilet paper

rhydex 247: here is the question again
Show that V=IR^2 is not a vector space over IR with respect to operation (a,b)+(c,d)=(a+c, b+d) and k(a,b)=(ka,kb)

If k is a real number, then the plane IR² is a vector space over the scalar field IR with respect to addition and scalar multiplication. IR² is then said to be a real vector space since the underlying scalar field is IR, the set of real numbers.

This question doesn't make sense. It is like asking you to prove that black is equal to white. There's nothing to prove really. Tear that question or use eraser wipe the "not" in the question.

@DoubleDx, check Page 81. I posted my solution on that question. I used a totally different elementary approach. You will see the upload on my signature. It is a pdf file.

Download it. I need your comment on the solution.