# 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

May 29, 2018

${s}_{n} = \frac{1}{12} n \left(n + 1\right) \left(n + 2\right) \left(3 n + 5\right)$

#### Explanation:

${s}_{n} = {\sum}_{k = 1}^{n} k {\left(k + 1\right)}^{2} = {\sum}_{k = 2}^{n + 1} \left(k - 1\right) {k}^{2}$

${s}_{n} = {\sum}_{k = 2}^{n + 1} {k}^{3} - {\sum}_{k = 2}^{n + 1} {k}^{2}$

${s}_{n} = {\sum}_{k = 1}^{n + 1} {k}^{3} - {\sum}_{k = 1}^{n + 1} {k}^{2}$

I won't derive these common sums:

The sum of the squares is

${\sum}_{k = 1}^{N} {k}^{2} = \frac{1}{6} N \left(N + 1\right) \left(2 N + 1\right)$

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

${\sum}_{k = 1}^{N} {k}^{3} = {\left(\frac{1}{2} N \left(N + 1\right)\right)}^{2}$

Now,

${s}_{n} = {\sum}_{k = 1}^{n + 1} {k}^{3} - {\sum}_{k = 1}^{n + 1} {k}^{2}$

$= {\left(\frac{1}{2} \left(n + 1\right) \left(n + 2\right)\right)}^{2} - \frac{1}{6} \left(n + 1\right) \left(n + 2\right) \left(2 \left(n + 1\right) + 1\right)$

$= \frac{1}{4} {\left(n + 1\right)}^{2} {\left(n + 2\right)}^{2} - \frac{1}{6} \left(n + 1\right) \left(n + 2\right) \left(2 n + 3\right)$

$= \frac{1}{12} \left(3 {\left(n + 1\right)}^{2} {\left(n + 2\right)}^{2} - 2 \left(n + 1\right) \left(n + 2\right) \left(2 n + 3\right)\right)$

$= \frac{1}{12} \left(n + 1\right) \left(n + 2\right) \left(3 \left(n + 1\right) \left(n + 2\right) - 2 \left(2 n + 3\right)\right)$

$= \frac{1}{12} \left(n + 1\right) \left(n + 2\right) \left(3 {n}^{2} + 9 n + 6 - 4 n - 6\right)$

${s}_{n} = \frac{1}{12} n \left(n + 1\right) \left(n + 2\right) \left(3 n + 5\right)$

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 $i$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

${S}_{n} = \frac{1}{3} {\sum}_{i = 1}^{n} {i}^{3} + \frac{1}{2} {\sum}_{i = 1}^{n} {i}^{2} + \frac{1}{6} {\sum}_{i = 1}^{n} i$

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

${\sum}_{i = 1}^{n} {i}^{3} = {\left(\frac{1}{2} n \left(n + 1\right)\right)}^{2}$

The triangular numbers are the sums of the naturals,

${\sum}_{i = 1}^{n} i = \frac{1}{2} n \left(n + 1\right)$

${S}_{n} = \frac{1}{3} {\left(\frac{1}{2} n \left(n + 1\right)\right)}^{2} + \frac{1}{2} \left(\frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right)\right) + \frac{1}{6} \left(\frac{1}{2} n \left(n + 1\right)\right)$

${S}_{n} = \frac{1}{12} n {\left(n + 1\right)}^{2} \left(n + 2\right)$