Question

If S₁, S₂, ..., S_n are the sums of n terms of n G.P.'s whose first term is 1 in each and common ratios are 1, 2, 3, ..., n respectively, then prove that S₁ + S₂ + 2S₃ + 3S₄ + ... (n − 1) S_n = 1ⁿ + 2ⁿ + 3ⁿ + ... + nⁿ.

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ {S}_1,S_2,...,S_n\mathrm{are}\mathrm{the}\mathrm{sum}\mathrm{of}n\mathrm{terms}\mathrm{of}\mathrm{an}\mathrm{G}.\mathrm{P}.\mathrm{whose}\mathrm{first}\mathrm{term}\mathrm{is}1\mathrm{in}\mathrm{each}\mathrm{case}\mathrm{and}\mathrm{the}\mathrm{common}\mathrm{ratios}\mathrm{are}1,2,3,...,n. \\ \therefore S_1=1+1+1+...n\mathrm{terms}=n...\left(1\right) \\ {S}_2=\frac{1\left(2^n-1\right)}{2-1}=2^n-1...\left(2\right) \\ {S}_3=\frac{1\left(3^n-1\right)}{3-1}=\frac{3^n-1}{2}...\left(3\right) \\ {S}_4=\frac{1\left(4^n-1\right)}{4-1}=\frac{4^n-1}{3}...\left(4\right) \\ . \\ . \\ . \\ . \\ {S}_n=\frac{1\left(n^n-1\right)}{n-1}=\frac{n^n-1}{n-1}............\left(\mathrm{n}\right) \\ \mathrm{Now},\mathrm{LHS}=S_1+S_2+2S_3+3S_4+...+\left(n-1\right)S_n \\ =n+2^n-1+3^n-1+4^n-1+...+n^n-1\left[\mathrm{Using}\left(1\right),\left(2\right),\left(3\right),...,\left(n\right)\right] \\ =n+\left(2^n+3^n+4^n+...+n^n\right)-\left[1+1+1+...+\left(n-1\right)\mathrm{times}\right] \\ =n+\left(2^n+3^n+4^n+...+n^n\right)-\left(n-1\right) \\ =n+\left(2^n+3^n+4^n+...+n^n\right)-n+1 \\ =1+2^n+3^n+4^n+...+n^n \\ =1^n+2^n+3^n+4^n+...+n^n \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$