71I think It's correct!!
11 Answers
71Well, Probably because When n --> ∞ the numbers becomes 0.
1but then it will be like this ::
x2 {11-x - xn1-x}
n->∞
so answer should be zero
1there 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|
71My 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|
1Actually 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)
1further :
x21-x + x 1-x1-x + log (1-x)
= x2 + x - x21-x + log (1-x)
= x1-x + log (1-x)
= ANSWER