How do we find the term n^(th) and sum to n^(th) of the following series?

1)1*2^2+2*3^2+3*4^2+...n^(th) term

2)1^2+(1^2+2^2)+(1^2+2^2+3^2)+....n^(th))term

2 Answers
May 29, 2018

s_n = 1/12 n (n + 1) (n + 2) (3 n + 5)

Explanation:

s_n = sum_{k=1}^n k (k+1)^2 = sum_{k=2}^{n+1} (k-1) k^2

s_n = sum_{k=2}^{n+1} k^3 - sum_{k=2}^{n+1} k^2

s_n = sum_{k=1}^{n+1} k^3 - sum_{k=1}^{n+1} k^2

I won't derive these common sums:

The sum of the squares is

sum_{k=1}^N k^2 = 1/6 N(N+1)(2N+1)

The sum of cubes is remarkably the square of the triangular numbers, the sums of the natural numbers.

sum_{k=1}^N k^3 = ( 1/2 N(N+1))^2

Now,

s_n = sum_{k=1}^{n+1} k^3 - sum_{k=1}^{n+1} k^2

= ( 1/2 (n+1)(n+2))^2 - 1/6 (n+1)(n+2)(2(n+1)+1)

= 1/4 (n+1)^2(n+2)^2-1/6(n+1)(n+2)(2n+3)

= 1/12 ( 3(n+1)^2(n+2)^2 - 2 (n+1)(n+2)(2n+3))

= 1/12 (n+1)(n+2)( 3(n+1)(n+2)-2(2n+3) )

= 1/12 (n+1)(n+2)( 3n^2+9n + 6 - 4n- 6)

s_n = 1/12 n (n + 1) (n + 2) (3 n + 5)

Is that a second sum? I'll leave that to others.

Jun 1, 2018

Second sum: sum_{i=1}^n sum_{j=1}^i j^2 = 1/12 n (n + 1)^2 (n + 2)

Explanation:

Let's do the second sum.

The ith term is sum_{j=1}^i j^2

There are n of those terms

S_n = sum_{i=1}^n sum_{j=1}^i j^2

The inner sum is known:

sum_{j=1}^i j^2 = 1/6 i (i+1)(2i+1) = i^3/3 + i^2/2 + i/6

S_n = 1/3 sum_{i=1}^n i^3 + 1/2 sum_{i=1}^n i^2 + 1/6 sum_{i=1}^n i

The sum of cubes is the square of the triangular number,

sum_{i=1}^n i^3 = ( 1/2 n(n+1))^2

The triangular numbers are the sums of the naturals,

sum_{i=1}^n i = 1/2 n(n+1)

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

S_n = 1/12 n (n + 1)^2 (n + 2)