Prove the following math problem...?

Question

The sum of k^2 (k is from 1 to n) equals 1/3 n^3 + 1/2 n^2 + 1/6 n

Answer 1

Prove it by induction

k =1

1/1 =1 and 1/3 +1/2+1/6 = 2/6 + 3/6 +1/6 = 1 too

Hypothese: k= n

Sum (1 to n) k^2 = (2n^2 + 3n+ 1)n/6 = (2n+1)(n+1)n/6

Thesis: k =n+1

Sum (1 to n+1) k^2 = (2n+3)(n+2)(n+1)/ 6

Proof

Sum (1 to n+1)k^2 = Sum(1 to kn k^2 + Sum(n+1 to n+1) k^2

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

= (n+1) [(2n+1)n+6(n+1)]/6 = (n+1)(2n^2+7n+6)/6=

= (n+1) (2n+3) (n+2)/6

Ana

Answer 2

This is a classic induction problem as other posters indicated. But there is another way, and this other way also answers the "who thought this stuff up anyway" question.

You set up a telescoping sum: sum((k+1)^3-k^3),k=1..n) which = [2^3-1^3] + [3^3-2^3] + [4^3-3^3] + ... +[n^3-(n-1)^3] + [(n+1)^3-n^3] = (n+1)^3 - 1 <<<<--- all the middle terms come in both + and - flavours and so cancel. On the other hand the polynomial (k+1)^3-k^3 inside the sum can be expanded: (k+1)^3-k^3 = (k^3+3k^2+3k+1) - k^3 = 3k^2+3k+1. So we get sum(3k^2+3k+1 = (n+1)^3-1 which becomes 3*sum(k^2) + 3*sum(k) + 3*sum(1) = (n+1)^3-1 then you substitute sum(k,k=1..n) = (n+1)*n/2 and sum(1,k=1..n) = n and solve for sum(k^2).

You could have used this trick with (k+1)^2-k^2 first to get sum(k), and after the step above, use it on (k+1)^4-k^4 to get sum(k^3) et ectera.

This method also shows that there is a formula for every integer power, e.g., there is some formula for sum(k^55,k=1..n).

Answer 3

...yet whilst 7n+4 is even, then n is atypical! you have have been given the question incorrect, i'm afraid... My suggestion for those questions like this could be to think of with reference to the way you will resolve the subject in the previous you start up writing something down -- What are you attempting to tutor? What effect are you searching for? and so on. And definite, announcing "2k=even" and "2k+a million is atypical" are in all probability the main common kit for one among those info.

Answer 4

This is a problem that lends itself to proof by induction.
1) Show that IF it is true for n then it is also true for n+1
2) Show that it is true for n=1

If you can show both then it must be true for n=1,2,3,4,...

Answer 5

I trhink you mean slove it .Its already proved I think .