Question

If $s_n$ denote the sum to $n$ terms of an A.P., then $s_{n+3}-3s_{n+2}+3s_{n+1}-s_n$ equals

• A. $1$
• B. $2$
• C. $0$
• D. $\frac{1}{3}$

Answer 1

The correct option is C $0$

$S_{n+3}-{3S}_{n+2}+3S_{n+1}-Sn$

$=\frac{n+3}{2}[2A+(n+3-1)D]-3\frac{n+2}{2}[2A+(n+2-1)D]$

$+3\frac{n+1}{2}[2A+(n+1-1)D]-\frac{n}{2}[2A-(n-1)D]$

$=A[(n+3)-3(n+2)-3(n+1)-n]$

$+\frac{D}{2}[(n+3)(n+2)-3(n+2)+(n+1)+3(n+1)n-n(n-1)]$

$=A[6-6]+\frac{D}{2}[6n-6n]=0$