Question

Sum of series $1+5+11+19+29+\dots \dots$ to $n$ terms.

• A. $S=\frac{n}{3}[n^2+3n-1]$
• B. $S=\frac{n}{3}[n^2+3n-2]$
• C. $S=\frac{2n}{3}[n^2+3n-1]$
• D. $S=\frac{n}{3}[n^2-3n-1]$

Answer 1

The correct option is A $S=\frac{n}{3}[n^2+3n-1]$

$\text{Solution: Let}S=1+5+11+19+29+\dots +a_n$

$S=1+5+11+19+\dots +a_{n-1}+a_n$

$0=1+4+6+8+10+\cdots +a_{n-1}-a_n$

$T_n=1+(4+6+8+10+\dots +a_{n-1}$

$\Rightarrow a=4;d=2$

$T_n=1+\frac{n-1}{2}[8+(n-2\left)2\right]$

$=1+\frac{n-1}{2}[8+2n-4]$

$=1+\frac{n-1}{2}[4+2n]$$

$=1+(n-1)(2+n)$

$=1+n^2+n-2$

$=n^2+n-1$

$S=\sum T_n=\sum n^2+\sum n-\sum 1$

$S=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}-n$

$S=n[\frac{(n+1)(2n+1)}{6}+\frac{n+1}{2}-1]$

$S=\frac{n}{6}\left[\right(n+1\left)\right(2n+1)+3(n+1)-6]$

$S=\frac{n}{6}[2n^2+3n+1+3n+3-6]$

$S=\frac{n}{6}[2n^2+6n-2]$

$S=\frac{n}{3}[n^2+3n-1]$