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Question: The sum of the series 1.2.3 + 2.3.4 + 3.4.5 + ….. to *n* terms is...

The sum of the series 1.2.3 + 2.3.4 + 3.4.5 + ….. to n terms is

A

n(n+1)(n+2)n ( n + 1 ) ( n + 2 )

B

(n+1)(n+2)(n+3)( n + 1 ) ( n + 2 ) ( n + 3 )

C

14n(n+1)(n+2)(n+3)\frac { 1 } { 4 } n ( n + 1 ) ( n + 2 ) ( n + 3 )

D

14(n+1)(n+2)(n+3)\frac { 1 } { 4 } ( n + 1 ) ( n + 2 ) ( n + 3 )

Answer

14n(n+1)(n+2)(n+3)\frac { 1 } { 4 } n ( n + 1 ) ( n + 2 ) ( n + 3 )

Explanation

Solution

Tn=n(n+1)(n+2)=n3+3n2+2nT _ { n } = n ( n + 1 ) ( n + 2 ) = n ^ { 3 } + 3 n ^ { 2 } + 2 n

∴ S=(n(n+1)2)2+3n(n+1)(2n+1)6+2n(n+1)2=14n(n+1)[n(n+1)+2(2n+1)+4]S = \left( \frac { n ( n + 1 ) } { 2 } \right) ^ { 2 } + 3 \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } + 2 \frac { n ( n + 1 ) } { 2 } = \frac { 1 } { 4 } n ( n + 1 ) [ n ( n + 1 ) + 2 ( 2 n + 1 ) + 4 ] =14n(n+1)[n2+5n+6]=14n(n+1)(n+2)(n+3)= \frac { 1 } { 4 } n ( n + 1 ) \left[ n ^ { 2 } + 5 n + 6 \right] = \frac { 1 } { 4 } n ( n + 1 ) ( n + 2 ) ( n + 3 )