## anonymous one year ago A = (1+2)/3 + (1+2+3)/3^2 + (1+2+3+4)/3^3 + ... The value of 8A = ....

1. anonymous

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

2. ganeshie8

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

3. imqwerty

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 :)

4. anonymous

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

5. imqwerty

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

6. anonymous

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 $\frac{1+2}{3}+\frac{1+2+3}{3^2}+\cdots=\sum_{n=1}^\infty \frac{\displaystyle\sum_{i=1}^{n+1}i}{3^n}=\sum_{n=1}^\infty \frac{(n+1)(n+2)}{2(3^n)}$

7. anonymous

(Minor typo in my first comment)

8. anonymous

@ganeshie8, how to massage it ? ._.