Question

Prove that for every positive integer n, $\sum ^n k{2}^k=(n-1)2^{n+1}+2$

Short answer

$\sum ^n k{2}^k=(n-1)2^{n+1}+2$

Step-by-step solution

Step: 1

Basic Step:

Now, we will show that P (1) is true.

$\begin{array}{r}\sum _{k=1}^n k{2}^k={1.2}^1=2\\ (1-1)2^{1+1}+2=2\end{array}$

Also,

Step: 2

Hence, as the both sides of P(1) are equal. So, P(1) is true.

Inductive hypotheses: As per this statement for all positive integer m, P(m) is true.

$\sum _{k=1}^n k{2}^k=(m-1)\cdot 2^{m+1}+2$

Step: 3

We need to show that P(m+1) is also true.

$\begin{array}{r}\sum _{k=1}^{m+1} k{2}^k=(m+1)\cdot 2^{m+1}+\sum _{k=1}^m k{2}^k\\ =(m+1)2^{m+1}+(m-1)2^{m+1}+2\\ =\left(2m\right)2^{m+1}+2\\ =m{2}^{m+2}+2\\ =(m+1)-{12}^{(m+1)+1}+2\end{array}$

So, from the above statement we can say that P (m+1) is also true.