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Question Number 155802 by cortano last updated on 05/Oct/21

 find the value of    2×1!+4×2!+6×3!+...+200×100!

$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\mathrm{2}×\mathrm{1}!+\mathrm{4}×\mathrm{2}!+\mathrm{6}×\mathrm{3}!+...+\mathrm{200}×\mathrm{100}! \\ $$

Answered by talminator2856791 last updated on 05/Oct/21

 use the information in the post    above this post Q.155803:   Σ_(k=1) ^n k×k! = (n+1)!−1    2×1!+4×2!+.....+200×100!   = 2(1+1!+2×2!+.....+100×100!)    = 2Σ_(k=1) ^(100) k×k!   = 2((100+1)!−1)   = determinant (((2×101!−2)))

$$\:\mathrm{use}\:\mathrm{the}\:\mathrm{information}\:\mathrm{in}\:\mathrm{the}\:\mathrm{post}\: \\ $$$$\:\mathrm{above}\:\mathrm{this}\:\mathrm{post}\:\mathrm{Q}.\mathrm{155803}: \\ $$$$\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}×{k}!\:=\:\left({n}+\mathrm{1}\right)!−\mathrm{1}\: \\ $$$$\:\mathrm{2}×\mathrm{1}!+\mathrm{4}×\mathrm{2}!+.....+\mathrm{200}×\mathrm{100}! \\ $$$$\:=\:\mathrm{2}\left(\mathrm{1}+\mathrm{1}!+\mathrm{2}×\mathrm{2}!+.....+\mathrm{100}×\mathrm{100}!\right)\: \\ $$$$\:=\:\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}{k}×{k}! \\ $$$$\:=\:\mathrm{2}\left(\left(\mathrm{100}+\mathrm{1}\right)!−\mathrm{1}\right) \\ $$$$\:=\begin{array}{|c|}{\mathrm{2}×\mathrm{101}!−\mathrm{2}}\\\hline\end{array} \\ $$

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