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Toolkit 74

Triangle Method for Identities with Coefficients

Start with:

1⋅21+2⋅22+3⋅23+⋯+(n−1)2n−1+n 2n.1\cdot 2^1+2\cdot 2^2+3\cdot 2^3+\cdots+(n-1)2^{n-1}+n\,2^n.

Write the following equations in a triangular arrangement:

21+22+23+⋯+2n−1+2n ⟶ 2n+1−21  22+23+⋯+2n−1+2n ⟶ 2n+1−22  23+⋯+2n−1+2n ⟶ 2n+1−23  ⋮  2n−1+2n ⟶ 2n+1−2n−1 2n ⟶ 2n+1−2n\begin{array}{l} 2^1+2^2+2^3+\cdots+2^{n-1}+2^n\ \longrightarrow\ 2^{n+1}-2^1\\[2pt] \qquad\ \ 2^2+2^3+\cdots+2^{n-1}+2^n\ \longrightarrow\ 2^{n+1}-2^2\\[2pt] \qquad\qquad\ \ 2^3+\cdots+2^{n-1}+2^n\ \longrightarrow\ 2^{n+1}-2^3\\[2pt] \qquad\qquad\qquad\qquad\ \ \vdots\\[2pt] \qquad\qquad\qquad\qquad\qquad\ \ 2^{n-1}+2^n\ \longrightarrow\ 2^{n+1}-2^{n-1}\\[2pt] \qquad\qquad\qquad\qquad\qquad\qquad\qquad\ 2^n\ \longrightarrow\ 2^{n+1}-2^n \end{array}

Then add them:

=n 2n+1−(21+22+23+⋯+2n−1+2n).=n\,2^{n+1}-\left(2^1+2^2+2^3+\cdots+2^{n-1}+2^n\right).

Since

21+22+⋯+2n−1+2n=2n+1−21,2^1+2^2+\cdots+2^{n-1}+2^n=2^{n+1}-2^1,

we get

=n 2n+1−(2n+1−21)=n\,2^{n+1}-(2^{n+1}-2^1)
=(n−1)2n+1+21.=\boxed{(n-1)2^{n+1}+2^1}.
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