MI # 1

Prove that the sum of n terms of

1*1! + 2*2! + 3.3! + 4*4!+.....n*(n+1)!

is (n+1)!-1.

You wrote n=1 instead of n

Let f(n)donate the series (|Last term is n*n!).

The statement is true for some p.

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Prove that

k^4= (k^5)/5 + (k^4)/2 +(K^3)/3 -k/30

by mathematical induction for the next term (k+1)

When I do this all f-s between 2 and n reduct as you see because f(k) thakes part in f(k+1)-f(k) with sign "-" and in f(k)-f(k-1) whit sign "+".

So there left only -f(1) and +f(n+1):

Now it's not hard to prove that kk!= (k+1)!-k!:

(k+1)!-k!= ((k+1)k!)-k!=k!(k+1-1)=kk!

so for every k kk!= (k+1)!-k!=f(k+1)-f(k),where f(x)=x!

So the sum:

,

which have to be proven.]]>

I'll leave you to your maths teacher to teach you what's an integer.]]>

k! means the product of all positive integers, less or equal to k:

k!=1.2.3. ... .k]]>

so

1.1!+2.2!+3.3!+4.4!+...+(n).(n)!

Prove that the sum of n terms of

1*1! + 2*2! + 3.3! + 4*4!+.....n*(n+1)!

is (n+1)!-1.