Question

The $n^{th}$ term of a sequence of numbers is $a_n$ and given by the formula $a_n=a_{n-1}+2n$ for $n\ge 2$ and $a_1=1$.

Using the above information $a_n$ will be

• A. $a_n=n^2+n-1$
• B. $a_n=n^2+n+1$
• C. $a_n=n^2-n+1$
• D. $a_n=n^2-n-1$

Answer 1

The correct option is A $a_n=n^2+n-1$

$a_1=1a_2=a_1+4=1+4=5a_3=a_2+6=5+6=11a_4=a_3+8=11+8=19$

Hence
$S=1+5+11+19+\dots \dots +a_n\cdots \left(i\right)$
$S=1+5+11+\dots \dots +a_{n-1}+a_n\cdots \left(ii\right)$
$\overline{0=1+4+6+8+\dots \dots +(a_n-a_{n-1})-a_n}$
$a_n=1+\underset{(n-1)\text{terms}}{\underset{}{2(2+3+4+\dots \dots +(a_n-a_{n-1}\left)\right)}}$
$a_n=1+2.\frac{(n-1)}{2}[4+(n-2\left)\right]=1+(n-1)(n+2)=n^2+n-1$
$\text{sum}=\sum a_n=\sum n^2+\sum n-\sum 1=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}-n=\frac{n}{6}\left[\right(n+1\left)\right(2n+1)+3(n+1)-6]$
$=\frac{n}{6}[2n^2+3n+1+3n+3-6]=\frac{n}{6}[2n^2+6n-2]=\frac{n(n^2+3n-1)}{3}$