Differentiate Me!

we have $\sum\limits_{i=1}^{\infty }{\frac{{{i}^{5}}}{{{2}^{i}}}}=\sum\limits_{i=0}^{\infty }{\frac{{{\left( i+1 \right)}^{5}}}{{{2}^{i+1}}}=\frac{1}{2}\sum\limits_{i=0}^{\infty }{\frac{{{\left( i+1 \right)}^{5}}}{{{2}^{i}}}}}$

now by using the binomial expand for ${{\left( i+1 \right)}^{5}}={{i}^{5}}+5{{i}^{4}}+10{{i}^{3}}+10{{i}^{2}}+5i+1$

So $\sum\limits_{i=1}^{\infty }{\frac{{{i}^{5}}}{{{2}^{i}}}=\frac{1}{2}\left( \sum\limits_{i=0}^{\infty }{\frac{{{i}^{5}}}{{{2}^{i}}}+5\sum\limits_{i=0}^{\infty }{\frac{{{i}^{4}}}{{{2}^{i}}}+10\sum\limits_{i=0}^{\infty }{\frac{{{i}^{3}}}{{{2}^{i}}}+10\sum\limits_{i=0}^{\infty }{\frac{{{i}^{2}}}{{{2}^{i}}}+5\sum\limits_{i=0}^{\infty }{\frac{i}{{{2}^{i}}}+\sum\limits_{i=0}^{\infty }{\frac{1}{{{2}^{i}}}}}}}}} \right)}$

But $\sum\limits_{i=0}^{\infty }{\frac{1}{{{2}^{i}}}=1+\sum\limits_{i=1}^{\infty }{{{\left( \frac{1}{2} \right)}^{i}}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=2}}$ & $\sum\limits_{i=0}^{\infty }{\frac{i}{{{2}^{i}}}=0+\sum\limits_{i=1}^{\infty }{\frac{i}{{{2}^{i}}}}}=2$

since $\sum\limits_{i=1}^{\infty }{\frac{i}{{{2}^{i}}}=\sum\limits_{i=0}^{\infty }{\frac{i+1}{{{2}^{i+1}}}=}\frac{1}{2}\left( \sum\limits_{i=0}^{\infty }{\frac{i}{{{2}^{i}}}+\sum\limits_{i=0}^{\infty }{\frac{1}{{{2}^{i}}}}} \right)=\frac{1}{2}\left( \sum\limits_{i=1}^{\infty }{\frac{i}{{{2}^{i}}}+2} \right)}$ $\Rightarrow \left( 1-\frac{1}{2} \right)\sum\limits_{i=1}^{\infty }{\frac{i}{{{2}^{i}}}=1}$ $\Rightarrow \sum\limits_{i=1}^{\infty }{\frac{i}{{{2}^{i}}}}=2$

Thus $\sum\limits_{i=1}^{\infty }{\frac{{{i}^{2}}}{{{2}^{i}}}}=\frac{1}{2}\left( \sum\limits_{i=1}^{\infty }{\frac{{{i}^{2}}}{{{2}^{i}}}+4+2} \right)=\frac{1}{2}\sum\limits_{i=1}^{\infty }{\frac{{{i}^{2}}}{{{2}^{i}}}+3}$ $\Rightarrow \left( 1-\frac{1}{2} \right)\sum\limits_{i=1}^{\infty }{\frac{{{i}^{2}}}{{{2}^{i}}}=3\Leftrightarrow \sum\limits_{i=1}^{\infty }{\frac{{{i}^{2}}}{{{2}^{i}}}=6}}$

Same work we will work for $\sum\limits_{i=0}^{\infty }{\frac{{{i}^{3}}}{{{2}^{i}}}}\,\,\And \,\,\sum\limits_{i=0}^{\infty }{\frac{{{i}^{4}}}{{{2}^{i}}}}$

$\sum\limits_{i=0}^{\infty }{\frac{{{i}^{3}}}{{{2}^{i}}}=0+\sum\limits_{i=1}^{\infty }{\frac{{{i}^{3}}}{{{2}^{i}}}=\sum\limits_{i=0}^{\infty }{\frac{{{\left( i+1 \right)}^{3}}}{{{2}^{i+1}}}=\frac{1}{2}\left( \sum\limits_{i=0}^{\infty }{\frac{{{i}^{3}}}{{{2}^{i}}}+3\sum\limits_{i=0}^{\infty }{\frac{{{i}^{2}}}{{{2}^{i}}}+3\sum\limits_{i=0}^{\infty }{\frac{i}{{{2}^{i}}}+\sum\limits_{i=0}^{\infty }{\frac{1}{{{2}^{i}}}}}}} \right)}}}$

$=\frac{1}{2}\left( \sum\limits_{i=1}^{\infty }{\frac{{{i}^{3}}}{{{2}^{i}}}+3\left( 6 \right)+3\left( 2 \right)+2} \right)$ $\Rightarrow \sum\limits_{i=1}^{\infty }{\frac{{{i}^{3}}}{{{2}^{i}}}}=\frac{1}{2}\sum\limits_{i=1}^{\infty }{\frac{{{i}^{3}}}{{{2}^{i}}}+13}$ $\Rightarrow \sum\limits_{i=0}^{\infty }{\frac{{{i}^{3}}}{{{2}^{i}}}=26}$

Similarly we expand $\sum\limits_{i=0}^{\infty }{\frac{{{i}^{4}}}{{{2}^{i}}}=0+\sum\limits_{i=1}^{\infty }{\frac{{{i}^{4}}}{{{2}^{i}}}=\sum\limits_{i=0}^{\infty }{\frac{{{\left( i+1 \right)}^{4}}}{{{2}^{i+1}}}}}}=\frac{1}{2}\sum\limits_{i=0}^{\infty }{\frac{{{\left( i+1 \right)}^{4}}}{{{2}^{i}}}}$

$\Rightarrow \sum\limits_{i=0}^{\infty }{\frac{{{i}^{4}}}{{{2}^{i}}}}=\frac{1}{2}\left( \sum\limits_{i=0}^{\infty }{\frac{{{i}^{4}}}{{{2}^{i}}}+4\sum\limits_{i=0}^{\infty }{\frac{{{i}^{3}}}{{{2}^{i}}}+6\sum\limits_{i=0}^{\infty }{\frac{{{i}^{2}}}{{{2}^{i}}}+4\sum\limits_{i=0}^{\infty }{\frac{i}{{{2}^{i}}}+\sum\limits_{i=0}^{\infty }{\frac{1}{{{2}^{i}}}}}}}} \right)$

$=\frac{1}{2}\sum\limits_{i=0}^{\infty }{\frac{{{i}^{4}}}{{{2}^{i}}}+\frac{1}{2}\left( 4\left( 26 \right)+6\left( 6 \right)+4\left( 2 \right)+2 \right)}$ $\Rightarrow \sum\limits_{i=0}^{\infty }{\frac{{{i}^{4}}}{{{2}^{i}}}}=\frac{1}{2}\sum\limits_{i=0}^{\infty }{\frac{{{i}^{4}}}{{{2}^{i}}}+75}$ $\Leftrightarrow \sum\limits_{i=0}^{\infty }{\frac{{{i}^{4}}}{{{2}^{i}}}=150}$

Thus $\sum\limits_{i=1}^{\infty }{\frac{{{i}^{5}}}{{{2}^{i}}}=\frac{1}{2}\sum\limits_{i=0}^{\infty }{\frac{{{i}^{5}}}{{{2}^{i}}}+\frac{1}{2}\left( 5\left( 150 \right)+10\left( 26 \right)+10\left( 6 \right)+5\left( 2 \right)+2 \right)}}$ thus $\sum\limits_{i=1}^{\infty }{\frac{{{i}^{5}}}{{{2}^{i}}}=1082}$ hence $\sqrt{A+7}=\sqrt{1082+7}=\sqrt{1089}=\sqrt{{33}^{2}}=33$

We all know that $\sum\limits _{ n=1 }^{ \infty }{ { x }^{ n } } =\frac { x }{ 1-x }$ for $\left| x \right| <1$ .

So, $\frac { d }{ dx } \left( \sum\limits _{ n=1 }^{ \infty }{ { x }^{ n } } \right) =\sum\limits _{ n=1 }^{ \infty }{ \frac { d }{ dx } { \left( { x }^{ n } \right) } } =\sum\limits _{ n=1 }^{ \infty }{ n{ x }^{ n-1 } } =\frac { d }{ dx } \left( \frac { x }{ 1-x } \right)\Rightarrow \sum\limits _{ n=1 }^{ \infty }{ n{ x }^{ n } } =x\frac { d }{ dx } \left( \frac { x }{ 1-x } \right)$

If this process is done 5 times, we get: $\sum\limits _{ n=1 }^{ \infty }{ { n }^{ 5 }{ x }^{ n } } = x\frac { d }{ dx } \left( x\frac { d }{ dx } \left( x\frac { d }{ dx } \left( x\frac { d }{ dx } \left( x\frac { d }{ dx } \left( \frac { x }{ 1-x } \right) \right) \right) \right) \right)$

So for $x=\frac { 1 }{ 2 }$ , $\sum\limits _{ n=1 }^{ \infty }{ \frac { { n }^{ 5 } }{ { 2 }^{ n } } } = { \left[ x\frac { d }{ dx } \left( x\frac { d }{ dx } \left( x\frac { d }{ dx } \left( x\frac { d }{ dx } \left( x\frac { d }{ dx } \left( \frac { x }{ 1-x } \right) \right) \right) \right) \right) \right] }_{ x=\frac { 1 }{ 2 } }$

After some calculation, you get $\sum\limits _{ n=1 }^{ \infty }{ \frac { { n }^{ 5 } }{ { 2 }^{ n } } } ={ \left[ \frac { x(x^{ 4 }+26x^{ 3 }+66x^{ 2 }+26x+1) }{ (x-1)^{ 6 } } \right] }_{ x=\frac { 1 }{ 2 } } =1082$ and finally $\sqrt { 1082+7 } =33$

The solution below is not written by me , i just share it from another forum which i asked.

Apparently it uses something called Poyla form.

The solver is the forum user Hoempa , and this is the link

As $\large \displaystyle \sum_{n=1}^{\infty} \frac{{n \choose k}}{2^n} = 2$

and $n^5 = 120{n \choose 5}+240{n \choose 4}+150{n \choose 3}+30{n \choose 2}+1{n \choose 1}$ , $\sum_{n=1}^{\infty} \frac{n^5}{2^n} = 2(120+240+150+30+1)=1082$

therefore $\sqrt{A+7}=\sqrt{1089}=33$

I posted here only after i obtained permission from the solver. I hope i credit it enough to not being called plagiariser.

Lastly , i don't understand this solution , never learn Poyla before. Please don't ask me anything significant. Just sharing for maybe.......the enlightment of some other user?

Let $S = \displaystyle \sum_{i=1}^\infty \frac{i^{5}}{2^{i}}$ .

Thus, $S = \frac{1^{5}}{2^{1}} + \frac{2^{5}}{2^{2}} + \frac{3^{5}}{2^{3}} + \ldots + \frac{n^{5}}{2^{n}} + \ldots$

Then, multiplying it by $2$ gives us: $2S = \frac{1^{5}}{2^{0}} + \frac{2^{5}}{2^{1}} + \frac{3^{5}}{2^{2}} + \ldots + \frac{(n + 1)^{5}}{2^{n}} + \ldots$

Thus, we can write: $2S - S = 1 + \frac{2^{5} - 1^{5}}{2^{1}} + \ldots + \frac{(n + 1)^5 - n^{5}}{2^{n}} + \ldots$

So: $S = 1 + 5\displaystyle \sum_{i=1}^\infty \frac{i^{4}}{2^{i}} + 10\displaystyle \sum_{i=1}^\infty \frac{i^{3}}{2^{i}} + 10\displaystyle \sum_{i=1}^\infty \frac{i^{2}}{2^{i}} + 5\displaystyle \sum_{i=1}^\infty \frac{i}{2^{i}} + \displaystyle \sum_{i=1}^\infty \frac{1}{2^{i}}$

I will calculate this by parts.

First, let $S_{1} = \displaystyle \sum_{i=1}^\infty \frac{1}{2^{i}}$ ; this is a geometric series with first term equal to $\frac{1}{2}$ and ratio $\frac{1}{2}$ ; thus, its value equals $\frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1$ .

Now, let $S_{2} = \displaystyle \sum_{i=1}^\infty \frac{i}{2^{i}}$ . We can write:

$S_{2} = \frac{1}{2^{1}} + \frac{2}{2^{2}} + \frac{3}{2^{3}} + \ldots + \frac{n}{2^{n}} + \ldots$

Multiplying it by two yields:

$2S_{2} = 1 + \frac{2}{2^{1}} + \frac{3}{2^{2}} + \frac{4}{2^{3}} + \ldots + \frac{(n + 1)}{2^{n}} + \ldots$

Thus: $2S_{2} - S_{2} = 1 + \frac{1}{2^{1}} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \ldots + \frac{1}{2^{n}} + \ldots$ .

From this we derive: $S_{2} = 1 + S_{1} = 1 + 1 = 2$ .

Thirdly, let $S_{3} = \displaystyle \sum_{i=1}^\infty \frac{i^{2}}{2^{i}}$ . We can write:

$S_{3} = \frac{1^{2}}{2^{1}} + \frac{2^{2}}{2^{2}} + \frac{3^{2}}{2^{3}} + \ldots + \frac{n^{2}}{2^{n}} + \ldots$

Multiplying it by two yields:

$2S_{3} = 1 + \frac{2^{2}}{2^{1}} + \frac{3^{2}}{2^{2}} + \frac{4^{2}}{2^{3}} + \ldots + \frac{(n + 1)^{2}}{2^{n}} + \ldots$

Thus: $2S_{3} - S_{3} = 1 + \frac{3}{2^{1}} + \frac{5}{2^{2}} + \frac{7}{2^{3}} + \ldots + \frac{(2n + 1)}{2^{n}} + \ldots$ .

From this we derive: $S_{2} = 1 + 2*S_{2} + S_{1} = 1 + 4 + 1 = 6$

Now, for $S_{4} = \frac{1^{3}}{2^{1}} + \frac{2^{3}}{2^{2}} + \frac{3^{3}}{2^{3}} + \ldots + \frac{n^{3}}{2^{n}} + \ldots$

Using a similar procedure we arrive at: $S_{4} = 1 + 3S_{3} + 3S_{2} + S_{1} = 1 + 3*6 + 3*2 + 1 = 26$

Finally, for $S_{5} = \frac{1^{4}}{2^{1}} + \frac{2^{4}}{2^{2}} + \frac{3^{4}}{2^{3}} + \ldots + \frac{n^{4}}{2^{n}} + \ldots$ , we have:

$S_{5} = 1 + 4S_{4} + 6S_{3} + 4S_{2} + S_{1} =$

$1 + 4*26 + 6*6 + 4*2 + 1 = 1 + 104 + 36 + 8 + 1 = 150$ .

Thus: $S = 1 + 5S_{5} + 10S_{4} + 10S_{3} + 5S_{2} + 1 =$

$1 + 5*150 + 10*26 + 10*6 + 5*2 + 1 = 12 + 750 + 260 + 60 = 1082$

Thus, $A = 1082$ , and so $\sqrt{A + 7} = \sqrt{1089} = 33$ .

For $-1< x < 1$ , we have:

$\begin{aligned} \sum_{n=0}^\infty x^n & = \frac 1{1-x} & \small {\color{#3D99F6}\text{Differentiate both sides}} \\ \sum_{n=1}^\infty nx^{n-1} & = \frac 1{(1-x)^2} & \small {\color{#3D99F6}\text{Multiply both sides by }x} \\ \sum_{n=1}^\infty nx^n & = \frac x{(1-x)^2} & \small {\color{#3D99F6}\text{Differentiate again}} \\ \sum_{n=1}^\infty n^2x^{n-1} & = \frac {1+x}{(1-x)^3} & \small {\color{#3D99F6} \times x \text{ again}} \\ \sum_{n=1}^\infty n^2x^n & = \frac {x(1+x)}{(1-x)^3} & \small {\color{#3D99F6}\text{Differentiate again}} \\ \sum_{n=1}^\infty n^3x^{n-1} & = \frac {1+4x+x^2}{(1-x)^4} & \small {\color{#3D99F6} \times x \text{ again}} \\ \sum_{n=1}^\infty n^3x^n & = \frac {x+4x^2+x^3}{(1-x)^4} & \small {\color{#3D99F6}\text{Differentiate again}} \\ \sum_{n=1}^\infty n^4x^{n-1} & = \frac {1+11x+11x^2+x^3}{(1-x)^5} & \small {\color{#3D99F6} \times x \text{ again}} \\ \sum_{n=1}^\infty n^4x^n & = \frac {x+11x^2+11x^3+x^4}{(1-x)^5} & \small {\color{#3D99F6}\text{Differentiate again}} \\ \sum_{n=1}^\infty n^5x^{n-1} & = \frac {1+26x+66x^2+26x^3+x^4}{(1-x)^6} & \small {\color{#3D99F6} \times x \text{ again}} \\ \sum_{n=1}^\infty n^5x^n & = \frac {x+26x^2+66x^3+26x^4+x^5}{(1-x)^6} & \small {\color{#3D99F6} \text{Put }x = \frac 12} \\ \sum_{n=1}^\infty \frac {n^5}{2^n} & = \frac {\frac 12 +\frac {26}{2^2}+\frac {66}{2^3}+\frac {26}{2^4}+\frac 1{2^5}}{\left(1-\frac 12\right)^6} \\ & = 2+26(4)+66(8) + 26(16) + 32 \\ & = 1082 \end{aligned}$

$\implies \sqrt{A+7} = \sqrt{1082+7} = \boxed{33}$ .

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For me I used differentiation. Just requires a lot of patience...