1!+2(2!)+3(3!)+...+n(n!)=(n+1)!-1

Mathematics

1 answer:

anyanavicka [17]2 years ago

7 0

Step-by-step explanation:

OK, let's assume it this way:

Sn=1.1!+2.2!+3.3!+...+n.n!=(2‐1).1!+(3-1).2!+(4-1)3!+...+((n+1)-1).n!

Sn=1.1!+2.2!+3.3!+...+n.n!=(2‐1).1!+(3-1).2!+(4-1)3!+...+((n+1)-1).n!=(2.1!-1!)+(3.2!-2!)+(4.3!-3!)+...+((n-1)n!-n!)=(2!-1!)+(3!-2!)+(4!-3!)+

Sn=1.1!+2.2!+3.3!+...+n.n!=(2‐1).1!+(3-1).2!+(4-1)3!+...+((n+1)-1).n!=(2.1!-1!)+(3.2!-2!)+(4.3!-3!)+...+((n-1)n!-n!)=(2!-1!)+(3!-2!)+(4!-3!)+...+(n+1)!-n!=(n+1)!-1!=(n+1)!-1

and boom problem solved

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1!+2(2!)+3(3!)+...+n(n!)=(n+1)!-1​

1!+2(2!)+3(3!)+...+n(n!)=(n+1)!-1