Question

If a${}_1$ = 4 and $a_{n+1}=a_n+4nforn\ge 1.$ then the value of a${}_{100}$ is

• A. 19804
• B. 18904
• C. 18894
• D. 19904
• E. 19894

Answer 1

Answer: (A)

Given, $a_{n+1}=a_n+4n$ and $a_1=4$
$\Rightarrow a_{n+1}-a_n=4n$
Put $n=1,a_2-a_1=4\cdot 1$
Put $n=2a_3-a_2=4\cdot 2$
Put $n=3,a_4-a_3=4\cdot 3$
$\dots \dots \dots \dots \dots \dots$
$\dots \dots \dots \dots \dots \dots$
Put $n=99,a_{100}-a_{99}=4\cdot 99$
On adding, we get
$a_{100}-a_1=4(1+2+\dots +99)$
$\Rightarrow a_{100}-4=4\cdot \frac{99(99+1)}{2}$
$\left[\because \sum n=\frac{n(n+1)}{2}\right]$
$\therefore a_{100}=19804$