Question

The $n^{\text{th}}$ 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 $a_1=1.$
The sum of first $20$ terms is

• A. $3060$
• B. $3059$
• C. $2680$
• D. $2679$

Answer 1

The correct option is A $3060$

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

Now,
$S=1+5+11+19+\dots \dots +a_n$
$S=1+5+11+\dots \dots +a_{n-1}+a_n$
$\overline{0=1+4+6+8+\dots \dots +(a_n-a_{n-1})-a_n}$
$a_n=1+\underset{(n-1)\text{terms}}{\underset{}{4+6+8+\dots \dots +(a_n-a_{n-1})}}$

$a_n=1+2\cdot \frac{(n-1)}{2}[4+(n-2\left)\right]\Rightarrow a_n=1+(n-1)(n+2)\Rightarrow a_n=n^2+n-1\Rightarrow a_n=n(n+1)-1S_n=\sum _{r=1}^n\left[r\right(r+1)-1]\Rightarrow S_n=\sum _{r=1}^{n}r(r+1)-\sum _{r=1}^{n}1\Rightarrow S_n=\frac{n(n+1)(n+2)}{3}-n\Rightarrow S_{20}=\frac{20\left(21\right)\left(22\right)}{3}-20\therefore S_{20}=3060$