prove this conjecture by mathematical induction?

Question

the sum as i goes from 1 to n of (n x n+1) = n(n+1)(n+2)/3
this is just so confusing!! every time i think i have it figured out i get so lost. any help would be great. 10 points to whoever completes the entire proof.
thanks for all your help

Answer 1

1) First, show it is true for n = 1:

∑(1 to 1) n*(n+1) = 1*2 = 2
and when n=1, n(n+1)(n+2)/3 = 1*2*3/3 = 2
So it is true for n=1.

2) Assume it is true for n = k
That is,
∑(1 to k) n*(n+1) = k(k+1)(k+2) / 3

3) Prove it is true for n = k+1

∑(1 to k+1) n*(n+1) = ∑[(1 to k) n*(n+1)] + (k+1)(k+2)
∑(1 to k+1) n*(n+1) = k(k+1)(k+2) / 3 + (k+1)(k+2)
∑(1 to k+1) n*(n+1) = (k+1)(k+2)*[k/3 + 1]
∑(1 to k+1) n*(n+1) = (k+1)(k+2)*(k+3)/3
or
∑(1 to k+1) n*(n+1) = (k+1) * ((k+1)+1) * ((k+1)+2) / 3

So it is true for n = k+1
Q.E.D.

Answer 2

first prove that the formula is true when n =1

1 x 2 = (1 x 2 x 3 ) / 3

Now prove that when n is true, then it is also true for n+1

sum(n+1) = n(n+1)(n+2)/3 + (n+1)(n+2) the next number

= [ n(n+1)(n+2) + 3(n+1)(n+2) ] /3
= [(n+1)( n(n+2) + 3(n+2) ] / 3
= [(n+1)( n^2 + 5n + 6)] / 3
= [(n+1)(n+2)(n+3)] / 3
=(n+1)( (n+1) + 1)((n+1) + 2)/3

so it is also true for n+1

Answer 3

For the Fibonacci series, shot that, for all n>=2, f(n)2 + f(n-a million)2 = f(2*n-a million). For the Fibonacci series, demonstrate that, for all n>2, f(n-a million)*f(n+a million) = f(n)2+(-a million)n. F(a million)=a million f(2)=a million f(3)=2 remember the backside case on your first evidence enable n=2. Then f(2)*2+f(a million)*2 = f(3) yet (a million)(2) + (a million)(2) = 4, not f(3)=2 i don't evaluate your notation could desire to be very sparkling. F(n)2 is seen to me as 2*f(n) f(a million), f(2), f(3) happy for all then you confirm for f(ok+a million) and it fairly is immediately proved for f(ok) won it