Question

If $S_n$ is the sum of the first n even natural numbers, then which of the following leave no remainder on dividing $S_n$ ?

• A. n
• B. n+1
• C. n and (n+1) both
• D. n-1

Answer 1

The correct option is C

n and (n+1) both

Since the sequence is a series of ‘n’ even numbers, the sequence is 2,4,6,8,…

The common difference, d =$t_2$ – $t_1$ = 4 – 2 = 2.

The first term, a = 2

Sum to ‘n’ terms,

$\Rightarrow S_n$ = $\frac{n}{2}(2a+(n-1\left)d\right)$

$\Rightarrow S_n$ = $\frac{n}{2}(2\times 2+(n-1\left)2\right)$

$\Rightarrow S_n$ = $\frac{n}{2}(4+2n-2)$

$\Rightarrow S_n$ = $\frac{n}{2}(2n+2)$

$\Rightarrow S_n$ = $n(n+1)$

i.e, $S_n$ is divisible by n and n+1.