Solution to ∀n ≥ 1, Xn i=1 i(i!) = (n + 1)! − 1 - Sikademy
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∀n ≥ 1, Xn i=1 i(i!) = (n + 1)! − 1

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Here's the Solution to this Question

\displaystyle{\sum^n_{i=1} }i(i!)=(n+1)!-1


proof by induction:


for n=1:

\displaystyle{\sum^1_{i=1} }1\cdot(1!)=1

(1+1)!-1=1


let for n=k:


\displaystyle{\sum^k_{i=1} }i(i!)=(k+1)!-1


then for n=k+1:


\displaystyle{\sum^{k+1}_{i=1} }i(i!)=\displaystyle{\sum^{k}_{i=1} }i(i!)+(k+1)(k+1)!=(k+1)!-1+(k+1)(k+1)!=


=(k+2)(k+1)!-1=(k+2)!-1


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