Question

Evaluate:

\[\large{1^2. 2! + 2^2. 3! +... + n^2 n!}\]

Answer 1

@Mertsj
@hartnn
@UnkleRhaukus
@Callisto

Answer 2

\[\sum_{1}^{n}(n^2.(n+1)!)\]

Answer 3

This looks better.

Answer 4

\[\sum_{k=1}^nk^2(k+1)!\]

Answer 5

\[\sum_{1}^{n}(k^2.(k+1)!)\]
\[= \sum_{1}^{n}(k^2.(k+1)k (k-1)!)\]
\[=\sum_{1}^{n}(k^2.(k+1).(k-1)!\]

Answer 6

*k^3 (in last step

Answer 7

\[\sum_{1}^{n}(k^3.(k+1).(k-1)!)\]

Answer 8

\[\sum_{1}^{n}(k^4 .(k-1)!) + \sum_{1}^{n}(k^3 . (k-1)!)\]

Answer 9

No idea after that........... :(

Answer 10

I think you'll have to use the third one from the bottom in the list: http://en.wikipedia.org/wiki/Summation#Some_summations_involving_binomial_coefficients_and_factorials

\[\sum_{k=0}^nk\cdot k!=(n+1)!-1\]
Wolfram gives a result that looks a lot like it: http://www.wolframalpha.com/input/?i=sum+k%5E2%28k%2B1%29%21+from+k%3D1+to+n

Answer 11

But, here we have (k+1)!k^2

Answer 12

You'll have to do some rearranging. I'm not sure how to go about this myself...

Answer 13

Okk I will try

Answer 14

Thanks

Answer 15

No on?