#19 Hua Luo Geng.

Algebra Level 3

1 2 + 4 4 + 9 8 + 16 16 + + n 2 2 n + = ? \frac{1}{2}+\frac{4}{4} + \frac{9}{8} +\frac{16}{16}+ \cdots + \frac{n^2}{2^n}+ \cdots = \ ?

6 10 12 8

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Marco Brezzi
Aug 20, 2017

n = 0 x n = 1 1 x c c c c c c x < 1 \displaystyle\sum_{n=0}^{\infty}x^n=\dfrac{1}{1-x}\phantom{cccccc} |x|<1

Taking the derivative on both sides and multiplying by x x

n = 0 n x n 1 = 1 ( 1 x ) 2 \displaystyle\sum_{n=0}^{\infty}nx^{n-1}=\dfrac{1}{(1-x)^2}

n = 0 n x n = x ( 1 x ) 2 \displaystyle\sum_{n=0}^{\infty}nx^n=\dfrac{x}{(1-x)^2}

Taking the derivative and multiplying by x x again

n = 0 n 2 x n 1 = x + 1 ( 1 x ) 3 \displaystyle\sum_{n=0}^{\infty}n^2x^{n-1}=\dfrac{x+1}{(1-x)^3}

n = 0 n 2 x n = x ( x + 1 ) ( 1 x ) 3 \displaystyle\sum_{n=0}^{\infty}n^2x^n=\dfrac{x(x+1)}{(1-x)^3}

We now have

S ( x ) = n = 0 n 2 x n = x ( x + 1 ) ( 1 x ) 3 S(x)=\displaystyle\sum_{n=0}^{\infty}n^2x^n=\dfrac{x(x+1)}{(1-x)^3}

We have to find

S ( 1 2 ) = 1 2 ( 1 2 + 1 ) ( 1 1 2 ) 3 = 6 S\left(\dfrac{1}{2}\right)=\dfrac{\dfrac{1}{2}\left(\dfrac{1}{2}+1\right)}{\left(1-\dfrac{1}{2}\right)^3}=\boxed{6}

Amazing Solution!

Kelvin Hong - 3 years, 9 months ago
Zach Abueg
Aug 20, 2017

Let n = 1 n 2 2 n = S \displaystyle \sum_{n \ = \ 1}^{\infty} \frac {n^2}{2^n} = S .

Note that

S = n = 1 n 2 2 n = n = 0 n 2 2 n \displaystyle S = \sum_{n \ = \ 1}^{\infty} \frac {n^2}{2^n} = \sum_{n \ = \ 0}^{\infty} \frac {n^2}{2^n}

and

S = n = 1 n 2 2 n = n = 0 ( n + 1 ) 2 2 n + 1 \displaystyle S = \sum_{n \ = \ 1}^{\infty} \frac {n^2}{2^n} = \sum_{n \ = \ 0}^{\infty} \frac {(n + 1)^2}{2^{n + 1}}

Thus,

S = n = 0 n 2 2 n = n = 0 ( n + 1 ) 2 2 n + 1 \displaystyle S = \sum_{n \ = \ 0}^{\infty} \frac {n^2}{2^n} = \sum_{n \ = \ 0}^{\infty} \frac {(n + 1)^2}{2^{n + 1}}

Now let's do some rewriting:

S = 2 S S = n = 0 ( n + 1 ) 2 2 n n = 0 n 2 2 n = n = 0 2 n + 1 2 n = n = 0 2 n 2 n + n = 0 1 2 n = 2 n = 0 1 2 n + n = 0 1 2 n = 4 + 2 = 6 \displaystyle \begin{aligned} S & = 2S - S \\ & = \sum_{n \ = \ 0}^{\infty} \frac {(n + 1)^2}{2^n} - \sum_{n \ = \ 0}^{\infty} \frac {n^2}{2^n} \\ & = \sum_{n \ = \ 0}^{\infty} \frac {2n + 1}{2^n} \\ & = \sum_{n \ = \ 0}^{\infty} \frac {2n}{2^n} + \sum_{n \ = \ 0}^{\infty} \frac {1}{2^n} \\ & = 2\sum_{n \ = \ 0}^{\infty} \frac {1}{2^n} + \sum_{n \ = \ 0}^{\infty} \frac {1}{2^n} \\ & = 4 + 2 \\ & = \boxed{6} \end{aligned}

There is a typo in the third line from the bottom, in the first summation it should be m 2 m \dfrac{m}{2^m}

Marco Brezzi - 3 years, 9 months ago

Log in to reply

Note that 2 n = 0 1 2 n = 2 n = 0 n 2 n \displaystyle 2 \sum_{{\color{#D61F06}{n \ = \ 0}}}^{\infty} \frac{1}{2^n} = 2 \sum_{{\color{#D61F06}{n \ = \ 0}}}^{\infty} \frac{n}{2^n} .

Zach Abueg - 3 years, 9 months ago

Log in to reply

Yes, I know. I just thought it was a typo, my bad

Marco Brezzi - 3 years, 9 months ago

Log in to reply

@Marco Brezzi No worries! I gotchu :)

Zach Abueg - 3 years, 9 months ago
Chew-Seong Cheong
Aug 21, 2017

S = n = 1 n 2 2 n = n = 0 n 2 2 n = n = 0 ( n + 1 ) 2 2 ( n + 1 ) = n = 0 n 2 + 2 n + 1 2 2 n = 1 2 n = 0 n 2 2 n + 1 2 n = 1 n 2 n 1 + 1 2 n = 0 1 2 n Note that n = 1 n 2 n 1 is sum of AGP. = 1 2 S + 1 2 × 4 + 1 2 ( 1 1 1 2 ) See note. = 1 2 S + 2 + 1 1 2 S = 3 S = 6 \begin{aligned} S & = \sum_{\color{#3D99F6}n=1}^\infty \frac {n^2}{2^n} = \sum_{\color{#D61F06}n=0}^\infty \frac {n^2}{2^n} \\ & = \sum_{\color{#D61F06} n=0}^\infty \frac {(n+1)^2}{2^(n+1)} \\ & = \sum_{\color{#D61F06} n=0}^\infty \frac {n^2+2n+1}{2\cdot 2^n} \\ & = \frac 12 \sum_{\color{#D61F06} n=0}^\infty \frac {n^2}{2^n} + \frac 12{\color{#3D99F6}\sum_{n=1}^\infty \frac n{2^{n-1}}} + \frac 12 \sum_{\color{#D61F06} n=0}^\infty \frac 1{2^n} & \small \color{#3D99F6} \text{Note that } \sum_{n=1}^\infty \frac n{2^{n-1}} \text{ is sum of AGP.} \\ & = \frac 12 S + \frac 12 \times {\color{#3D99F6}4} + \frac 12 \left(\frac 1{1-\frac 12}\right) & \small \color{#3D99F6} \text{See note.} \\ & = \frac 12 S + 2 + 1 \\ \implies \frac 12 S & = 3 \\ S & = \boxed{6} \end{aligned}


Note: Arithmetic-geometric progression (AGP)

k = 1 n [ a + ( k 1 ) d ] r k 1 = a [ a + ( n 1 ) d ] r n 1 r + d r ( 1 r n 1 ) ( 1 r ) 2 \displaystyle \sum_{k=1}^n [a+(k-1)d]r^{k-1} = \frac {a-[a+(n-1)d]r^n}{1-r} + \frac {dr(1-r^{n-1})}{(1-r)^2}

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...