Question

A = (1+2)/3 + (1+2+3)/3^2 + (1+2+3+4)/3^3 + ...

The value of 8A = ....

Answer 1

The choices :
A. 3
B. 15
C. 19
D. 22

Answer 2

Recall the geometric series :
\[\sum\limits_{n=0}^{\infty}x^n = \dfrac{1}{1-x}\]

differentiate and get
\[\sum\limits_{n=1}^{\infty}nx^{n-1} = \dfrac{-1}{(1-x)^2}\]

differentiate again and get
\[\sum\limits_{n=2}^{\infty}n(n-1)x^{n-2} = \dfrac{2}{(1-x)^3}\]

plugin \(x=\dfrac{1}{3}\) and massage a bit

Answer 3

the whole sum A can be represented as -
\[\sum_{n=1}^{\infty}\frac{ (n+1)^2 }{ 2(3^n) }\]

can u evaluate this thing by breaking into parts :)

Answer 4

@imqwerty unless I'm missing something, did you mean
\[\sum_{n\ge1}\frac{n(n+1)}{2(3^n)}~~?\]

Answer 5

the numerator has n+1 terms with the 1st term=1 and common diff=1
so the numerator wuld be
(n+1)(2+(n-1))/2 =(n+1)(n+1)/2=(n+1)^2/2

Answer 6

Hmm, the reason I'm confused is that the numerators belong to the sequence of sums of consecutive positive integers, i.e.
\[\sum_{i=1}^ki=1+2+\cdots+k,\quad \text{for }k\ge2\]
This sum has a nice closed form
\[\sum_{i=1}^ki=\binom k2=\frac{k(k+1)}{2}\]
so the original series should be
\[\frac{1+2}{3}+\frac{1+2+3}{3^2}+\cdots=\sum_{n=1}^\infty \frac{\displaystyle\sum_{i=1}^{n+1}i}{3^n}=\sum_{n=1}^\infty \frac{(n+1)(n+2)}{2(3^n)}\]

Answer 7

(Minor typo in my first comment)

Answer 8

@ganeshie8, how to massage it ? ._.