Question

The sum to n terms of a series is given by $\frac { 1 } { 4 }$ n(n + 1) (n + 2) (n + 3) then n^th term of the series is –

• A. n(n + 1) (n + 2)
• B. n(4n² – 1)
• C. $\frac { \mathrm { n } ( \mathrm { n } + 1 ) ( 2 \mathrm { n } + 1 ) } { 6 }$
• D. (n + 1) (n + 2) (n + 3)

Answer 1

Answer: (A)

n(n + 1) (n + 2)

Answer 2

T_n = S_n – S_{n – 1}

= $\frac { 1 } { 4 }$(n – 1)

n (n + 1) (n + 2)

= n (n + 1) (n + 2).