How could we prove this ?∑i=1n(i2+3i+1)×i!=(n+3)×(n+1)!−3

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Drake Huang · Accepted Answer

We first look at the simpler expression∑k=1nk⋅k!This is1⋅1!+2⋅2!+3⋅3!+⋯+n⋅n!In general, k⋅k≠(k+1)⋅k!−k≠(k+1)!−k!. It follows that the sum above is equal to(1!−0!)+(2!−1!)+(3!−2!)+(4!−3!)+⋯+((n+1)!−n!)Add up. There is a whole lot of cancellation, and we get (n+1)!−1Now we turn to our problem, which can be rewritten as∑k=1n(k+1)2⋅k!+∑k=1nk⋅k!since k2+3k+1=(k+1)2+k. We have already computed the second sum. The first sum is∑k=1n(k+1)⋅(k+1)!The same collapsing argument then shows that this sum is(n+2)!−2Or else we can recycle the previous result, noting that we are dealing with ∑j=2n+1j⋅j!, which is 1 less than ∑j=1n+1j⋅j!. Finally, add up. We get[(n+2)(n+1)!−2]+[(n+1)!−1]This is (n+3)(n+1)!−3

allemelrypi0e · Accepted Answer

If you observe that i2+3i+1=(i+2)(i+1)−1, then your left-hand sum is∑i=1n((i+2)(i+1)−1)i!=∑i=1n((i+2)!−i!)As Henry noted, this sum telescopes into(n+2)!+(n+1)!−2!−1!=(n+2)(n+1)!+(n+1)!−3=(n+3)(n+1)!−3

How could we prove this ? \(\displaystyle{\sum_{{{i}={1}}}^{{n}}}{\left({i}^{{2}}+{3}{i}+{1}\right)}\times{i}!={\left({n}+{3}\right)}\times{\left({n}+{1}\right)}!-{3}\)

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