Question

Find the sum to $n$ terms of the series $5+11+19+29+41+...$

Answer 1

Step 1: Defining the problem :

The general form of the given series is,

$$\begin{gathered} t_n=n+{(n+1)}^2 \\ =n^2+3n+1 \end{gathered}$$

Where, $t_n$ is the $n$^th term.

Step 2: Summation using $\sum _{}$

Let $S_n$ be the sum of the series

$$\begin{gathered} \Rightarrow S_n=\sum _{}\left(n^2+3n+1\right) \\ \Rightarrow S_n=\sum _{}{n}^2+3\sum _{}n+\sum _{}1 \\ \Rightarrow S_n=\frac{n(n+1)(2n+1)}{6}+\frac{3n(n+1)}{2}+n \\ \Rightarrow S_n=\frac{n(2n^2+2n+n+1+9n+9+6)}{6} \\ \Rightarrow S_n=\frac{n(2n^2+12n+16)}{6} \\ \Rightarrow S_n=\frac{n(n^2+6n+8)}{3} \\ \Rightarrow S_n=\frac{n(n+4)(n+2)}{3} \end{gathered}$$

Therefore, $5+11+19+29+41+...$ up to $n$ is $\frac{n(n+4)(n+2)}{3}$