series

∞
Σ xn+1 = x2(1-x)
n=1

11 Answers

71
Vivek @ Born this Way ·

I think It's correct!!

1
prateek mehta ·

why

1
prateek mehta ·

plz tell ... why

i think the answr should be

x2 1- xn1-x

71
Vivek @ Born this Way ·

Well, Probably because When n --> ∞ the numbers becomes 0.

1
prateek mehta ·

but then it will be like this ::

x2 {11-x - xn1-x}

n->∞

so answer should be zero

1
prateek mehta ·

there can be another reason....

if we assume |x| < 1

then it is exactly as u say

....

because i have given only a part of question

this question also consists of another fraction added to this

∞
Σ xn+1n+1
n=1

so final answer is

x21-x + x + log|1-x|

71
Vivek @ Born this Way ·

My Condition is valid for -1 < x < 1, I think.
By the way, How did you obtain the last part of the Problem.. .+ x + log|1-x|

1
prateek mehta ·

Actually this is what the complete question/solution is :

x2 12 + x3 23 + x4 34 + x5 45 + .....

∞
Tn = Σ n xn+1n+1
n =1

∞
= Σ n + 1 - 1 xn+1n+1
n =1

= xn+1 { n+1n+1 - 1n+1 }

∞
= Σ xn+1 - xn+1n+1
n=1

= x2 1-xn1-x + log (1-x) + x

= x21-x + x + log (1-x)

1
prateek mehta ·

further :

x21-x + x 1-x1-x + log (1-x)

= x2 + x - x21-x + log (1-x)

= x1-x + log (1-x)

= ANSWER

1
prateek mehta ·

vivek ..do see this

1
prateek mehta ·

what are pink posts ? and how to get pink ?

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