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Prove that

$$\sum _{ i=1 }^{ n }{ \frac { i }{ { 2 }^{ i } } = } 2-\frac { n+2 }{ { 2 }^{ n } } \quad { for\quad all}\quad n\in \mathbb N $$

My attempt:

For my inductive step I tried the following:

$$\left(2 − \left(\frac{k + 2}{2^k}\right)\right) + \left(\frac{k + 1}{2^{k+1}}\right) = 2 − \left(\frac{k + 3}{2^{k+1}}\right)$$

But they never equal, this is where I'm stuck.

Shuri2060
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Heskinammo
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  • You need to show the LHS is equal to the RHS. A hint would be to use $2^{k+1}=2(2^k)$ – Shuri2060 Jul 11 '17 at 23:19
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    Assuming you meant $[2-(k+2)/2^k] + [(k+1)/2^{k+1}] = 2 - (k+3)/2^{k+1}$ in your last equation (you should probably brush up on TeX), I'm not sure what the problem is. This statement is true. – eyeballfrog Jul 11 '17 at 23:26
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    "But they never equal, this is where I'm stuck" what do you mean by they are never equal? As far as I can tell, your last equation is true for any $k$. – Simply Beautiful Art Jul 11 '17 at 23:45
  • Google for "StackExchange LaTeX guide". – user21820 Jul 12 '17 at 06:15

2 Answers2

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Given $ \sum _{ i=1 }^{ n }{ \frac { i }{ { 2 }^{ i } } = } 2-\frac { n+2 }{ { 2 }^{ n } } \quad for\quad all\quad n\in\mathbb N$

you should show

$$\sum _{ i=1 }^{ n+1 }{ \frac { i }{ { 2 }^{ i } } = } 2-\frac { n+3 }{ { 2 }^{ n+1 } } \quad $$

$$\sum _{ i=1 }^{ n+1 }{ \frac { i }{ { 2 }^{ i } } = } \sum _{ i=1 }^{ n }{ \frac { i }{ { 2 }^{ i } } +\frac { n+1 }{ { 2 }^{ n+1 } } } =2-\frac { n+2 }{ { 2 }^{ n } } +\frac { n+1 }{ { 2 }^{ n+1 } } =2+\frac { n+1-2n-4 }{ { 2 }^{ n+1 } } =2-\frac { n+3 }{ { 2 }^{ n+1 } } \\ $$

haqnatural
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hint

Let $$S_n=2-\frac {n+2}{2^n} . $$

then $$S_{n+1}-S_n=\frac {n+2}{2^n}-\frac {n+3}{2^{n+1}} $$

$$=\frac {2n+4-n-3}{2^{n+1}}=\frac {n+1}{2^{n+1}} $$

hamam_Abdallah
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