Question

Use mathematical induction to prove that the formula is true for every positive integer n.
1. 2+4+6+8+....+(2n)=n(n+1)
2. 1+4+7+10+.....+(3n-2)=n(3n-1)/2
3.1^3+2^3+3^3+4^3.....+(n^3)= (n^2(n+1)^2)/4
4. 3+7+11+15+.....+(4n-1)= n(2n+1)
5. 1/(1*2*3)+1/(2*3*4)+ 1/(3*4*5)+1(4*5*6)+.....+(1/(n(n+1)(n+2))=n(n+3)/4(n+1)(n+2))

Answer 1

Ok i would try the 3rd one for you so you will get the idea of it.

so we should prove 1^3+2^3+3^3+4^3.....+(n^3)= (n^2(n+1)^2)/4

first we should check if its correct for n=1
L.H.S.=1 R.H.S.=1(2^2)/4=1
so its correct for n=1.
assume it would be true when n=p
so 1^3+2^3+3^3+4^3.....+(p^3)= (p^2(p+1)^2)/4
then lets try to show that its true for n=p+1 too
like this
1^3+2^3+3^3+4^3.....+(p^3)= (p^2(p+1)^2)/4
adding (p+1)^3 for both sides
1^3+2^3+3^3+4^3.....+(p^3) +(p+1)^3= (p^2(p+1)^2)/4 +(p+1)^3
=(p+1)^2[p^2+4(p+1)]/4
=(p+1)^2[p^2+4p+4)]/4
=(p+1)^2[p+2^2)]/4
so it seems if we assume its correct for n=p it would be true to n=p+1 too.
and we showed that its correct for n=1 in the beginning.
so according to mathematical induction the result is true for every possible integer n

* its like a chain of domions

Answer 2

Is there any trick to reducing the equation to get it to look like the beginning one?

Answer 3

I can figure everything but the last step is hardest for me

Answer 4

Thnaks you mean reducing the equation? well i take common terms ( or what do u call them) out so the equation becomes simpler

Answer 5

oh ok thank you

Answer 6

you are welcome :)