Question

Sum of first $n$ natural numbers is $S_1$, sum of first $n$ odd natural numbers is $S_2$, and sum of first $n$ even numbers is $S_3$. Then $S_1:S_2:S_3$=

• A. $n-1:n:n+1$
• B. $n:2n-1-1:2n$
• C. $n+1:2n:2(n+1)$
• D. $n+1:n:2n-1:2n$

Answer 1

Answer: (C)

$n+1:2n:2(n+1)$

$S_n=\frac{n}{2}[2a+(n-1)d]$
$S_1=1+2+3+4+........n$
$a=1,d=1$
$S_1=\frac{n}{2}[2\left(1\right)+(n-1)1]$
$=\frac{n(n+1)}{2}$
$S_2=1+3+5+..........+n$
$a=1,d=2$
$S_2=\frac{n}{2}[2\left(1\right)+(n-1)2]$
$=\frac{n}{2}\left[2n\right]$
$=n^2$
$S_3=2+4+6+..........n$
$a=2,d=2$
$S_3=\frac{n}{2}[2\left(2\right)+(n-1)2]$
$=n[2+n-1]$
$=n(n+1)$
$S_1:S_2:S_3$
$\frac{n(n+1)}{2}:n^2:n(n+1)$
$\Rightarrow \frac{n+1}{2}:n:n+1$
$\Rightarrow (n+1):2n:2(n+1)$