# What is the value of the sum of the sequence: 1+5+14+30+........... upto n terms? Thank you!

##### 1 Answer
Nov 24, 2016

${s}_{n} = \frac{1}{12} \left({n}^{4} + 4 {n}^{3} + 5 {n}^{2} + 2 n\right)$

#### Explanation:

Notice that the sequence of differences between terms follows the pattern $4$, $9$, $16$, i.e. ${2}^{2}$, ${3}^{2}$, ${4}^{2}$.

Assuming the pattern continues, the next couple of terms would be $55$ and $91$, being $30 + {5}^{2}$ and $55 + {6}^{2}$

Write down the sequence of the first few sums:

$\textcolor{b l u e}{1} , 6 , 20 , 50 , 105 , 196$

Write down the sequence of differences between consecutive terms:

$\textcolor{b l u e}{5} , 14 , 30 , 55 , 91$

Write down the sequence of differences of that sequence:

$\textcolor{b l u e}{9} , 16 , 25 , 36$

Write down the sequence of differences of that sequence:

$\textcolor{b l u e}{7} , 9 , 11$

Write down the sequence of differences of that sequence:

$\textcolor{b l u e}{2} , 2$

Having reached a constant sequence, we can use the initial terms of each of these sequences as coefficients of an expression for the $n$th term of the sequence of sums:

s_n = color(blue)(1)/(0!) + color(blue)(5)/(1!)(n-1) + color(blue)(9)/(2!)(n-1)(n-2) + color(blue)(7)/(3!)(n-1)(n-2)(n-3) + color(blue)(2)/(4!)(n-1)(n-2)(n-3)(n-4)

$\textcolor{w h i t e}{{s}_{n}} = 1 + 5 \left(n - 1\right) + \frac{9}{2} \left({n}^{2} - 3 n + 2\right) + \frac{7}{6} \left({n}^{3} - 6 {n}^{2} + 11 n - 6\right) + \frac{1}{12} \left({n}^{4} - 10 {n}^{3} + 35 {n}^{2} - 50 n + 24\right)$

$\textcolor{w h i t e}{{s}_{n}} = \frac{1}{12} \left({n}^{4} + 4 {n}^{3} + 5 {n}^{2} + 2 n\right)$