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
Dean R.
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.

Dean R.
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 #i#th 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) #

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