*Question*

# Find the sum of n terms of the series \(3 + 8 + 22 + 72 + 266 + 1036 + … +\) up to n terms.

*Solution:*

Given \(S_n = 3 + 8 + 22 + 72 + 266 + 1036 + … +\) up to n terms

Their 2nd order difference is in GP : 9 , 36 , 144 , 576 , ….. with r = 4 and a = 9

*If any series has pth order differences in GP, then* \(f(n)\)* will represent its nth term where *\(f(n) = xr^n + [a \;polynomial\; of \;degree \;(p – 1)]\)

In this case 2nd order difference is in GP then we will use \(f(n) = xr^n + [a\; polynomial\; of\; degree\;(2 – 1)]\)

General term : \(T_n = xr^n + an + b\)

Then , \(T_1 = 4x + a + b = 3\)

\(T_2 = 16x + 2a + b = 8\)

\(T_3 = 64x + 3a + b = 22\)

Solving them we get \(x = \frac{1}{4} \: and \: a = 2 \: and \: b = 0\)

Now \(T_n = \frac{1}{4} \cdot 4^n + 2n\)

\(S_n = \sum_{n=1}^{n}T_n\)

\(\Rightarrow S_n = \sum_{n=1}{n}\frac{1}{4} \cdot 4^n + 2n\)

\(\Rightarrow S_n = \frac{1}{4}\sum_{n=1}{n}4^n + \sum_{n=1}{n}2n\)

\(\Rightarrow S_n = \frac{1}{4}(4 + 4^2 + 4^3 + … + 4^n) + \frac{2 \cdot n \cdot (n+1)}{2}\)

\(\Rightarrow S_n = \frac{1}{4}(\frac{4 \cdot (4^n – 1)}{4 – 1}) + n(n+1)\)

Therefore, \(\Rightarrow S_n = (\frac{4^n – 1}{3}) + n(n+1)\)