Solving Summations of n.n! Terms

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The discussion focuses on solving the summation of the series 1.1! + 2.2! + 3.3!... up to n terms. The first part of each term is identified as an arithmetic progression (AP) with a common difference of 1, and the nth term is expressed as n.n!. The participants explore the expansion of n.n! and consider substituting n with (n+1)-1 to simplify the computation. However, there is confusion regarding the application of this substitution, leading to requests for further clarification. The conversation emphasizes the need for clearer hints to effectively compute the summation.
utkarshakash
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Homework Statement


1.1!+2.2!+3.3!...n terms

Homework Equations



The Attempt at a Solution


The first part of every term is in AP whose cd is 1. Also the nth term of this series is given as n.n! If I expand it it becomes n.n(n-1)(n-2)...1

Now
S_n=\sum t_n \\<br /> = \sum n.n(n-1)(n-2)...1

But now how do I compute this summation?
 
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Consider n = (n+1)-1
 
Mentallic said:
Consider n = (n+1)-1

I didn't get you. What you said was to replace all the n's with (n+1)-1 but I didn't find it to be helpful. Can you please give me some more hints?
 
utkarshakash said:
I didn't get you. What you said was to replace all the n's with (n+1)-1 but I didn't find it to be helpful. Can you please give me some more hints?

Not all n's!

n.n! = [(n+1)-1]n!
 

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