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prove that 1(1^1)+2(2^2)+...n(n!)=(n+1)!-1 for all positive integers

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anditap | Student, Undergraduate

Posted October 23, 2011 at 11:15 AM via web

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prove that 1(1^1)+2(2^2)+...n(n!)=(n+1)!-1 for all positive integers

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jeyaram | Student, Undergraduate

Posted October 23, 2011 at 3:28 PM (Answer #1)

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

L.H.S= 1(1^1)+2(2^2)+...+n(n!)

n=1......L.H.S=1

n=1......R.H.S=(1+1)!-1

=2!-1

=2-1

=1

So when n=1 this is true

n=2...L.H.S=1(1^1)+2(2^2)

=5

n=2...R.H.S=(2+1)!-1

=3!-1

=6-1

=5

So when n=2 this is true

we take this is true when n=p (p is a real number)

so 1(1^1)+2(2^2)+...+p(p!)=(p+1)!-1 is true

n=(p+1)...L.H.S=1(1^1)+2(2^2)+...+p(p!)+(p+1)(p+1)!

put           1(1^1)+2(2^2)+...+p(p!)=(p+1)!-1

=(p+1)!-1+(p+1)(p+1)!

=(p+1)!+(p+1)(p+1)!-1

=(p+1)!{1+(p+1)}-1

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

=(p+2)!-1

so acording the math logical  1(1^1)+2(2^2)+...+n(n!)=(p+1)!-1 is true

 

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