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Question: The first term of an A.P. of consecutive integers is \(p ^ { 2 } + 1\) The sum of \(( 2 p + 1 )\) ...

The first term of an A.P. of consecutive integers is p2+1p ^ { 2 } + 1 The sum of (2p+1)( 2 p + 1 ) terms of this series can be expressed as.

A

(p+1)2( p + 1 ) ^ { 2 }

B

(p+1)3( p + 1 ) ^ { 3 }

C

(2p+1)(p+1)2( 2 p + 1 ) ( p + 1 ) ^ { 2 }

D

p3+(p+1)3p ^ { 3 } + ( p + 1 ) ^ { 3 }

Answer

p3+(p+1)3p ^ { 3 } + ( p + 1 ) ^ { 3 }

Explanation

Solution

S2p+1=2p+12{2(p2+1)+(2p+1−1)1}S _ { 2 p + 1 } = \frac { 2 p + 1 } { 2 } \left\{ 2 \left( p ^ { 2 } + 1 \right) + ( 2 p + 1 - 1 ) 1 \right\}

=(2p+12)(2p2+2p+2)=(2p+1)(p2+p+1)= \left( \frac { 2 p + 1 } { 2 } \right) \left( 2 p ^ { 2 } + 2 p + 2 \right) = ( 2 p + 1 ) \left( p ^ { 2 } + p + 1 \right)

=p3+(p+1)3= p ^ { 3 } + ( p + 1 ) ^ { 3 }.