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Series solution of $(2^k) \times 1 + (2^{k-1}) \times 2+ (2^{k-2}) \times 3+…+(2^2) \times(n-2) +2(n-1) +n?$
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Let $ \begin{align} S =(2^k) \cdot 1 + &(2^{k-1}) \cdot 2+ (2^{k-2}) \cdot 3+… +2(n-1) +n  \qquad \qquad \qquad \quad \rightarrow (i) \\ \frac{S}{2} = \qquad \qquad &(2^{k-1}) \cdot 1 + (2^{k-2}) \cdot 2+… +2(n-2) +(n-1) + \frac{n}{2}  \qquad \rightarrow (ii) \end{align} $

$eqn~(i)-(ii):$

$ \begin{align} \frac{S}{2} &= 2^k + 2^{k-1} + 2^{k-2} +… +2 +1 – \frac{n}{2} \\ \implies \frac{S}{2} &=\frac{2^{k+1} -1 }{2-1} – \frac{n}{2}  \\ \implies S &= 2\cdot(2^{k+1} – 1) – n = 2^{k+2} - n -2 \end{align}$
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