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After looking at the response, the formula would be $S_n = n2^{n+2} - (n+1)2^{n+1} + 2$

How would i use induction to prove this formula? I tried setting $S_{n+1}$ but having hard time getting $S_{n+1} = S_{n} + (n+1)2^{n+1}$

Edit: Also am i doing something wrong here: $S_{n} + (n+1)2^{n+1} = n2^{n+2} + 2.$ This look weird from $S_{n+1}$

Cup
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1 Answers1

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HINT

Let proceed by

  • base case: $S_1=1\cdot 2^{3} - (2)2^{2} + 2=2$
  • induction step: assume true $S_n=n2^{n+2} - (n+1)2^{n+1} + 2$ then we need to prove that

$$S_{n+1}=(n+1)2^{n+3} - (n+2)2^{n+2} + 2=\\=(n+1)2^{n+3} - (n+2)2^{n+2}-n2^{n+2} + (n+1)2^{n+1}+\overbrace{n2^{n+2} - (n+1)2^{n+1} + 2}^{S_n}\\\stackrel{?}=(n+1)2^{n+1}+S_n$$

that is

$$(n+1)2^{n+3} - (n+2)2^{n+2}-n2^{n+2} =0$$

user
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