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

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

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(p + 1)^{3} + 1

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(2 p_{1}) (p + 1)^{2} - 3

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None

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Solution

The correct option is A $p^{3} + (p + 1)^{3}$
$a = k^{2} + 1$
Since series is of consecutive integer
Sum of $(2 k + 1)$ terms $= \frac{n}{2} (2 a + (n - 1) d)$
$= \frac{(2 k + 1)}{2} [2 (k^{2} + 1) + (2 k + 1 - 1) 1]$
$= \frac{(2 k + 1)}{2} [2 k^{2} + 2 + 2 k]$
$= (2 k + 1) (k^{2} + k + 1)$
$= 2 k^{3} + 2 k^{2} + 2 k + k^{2} + k + 1$
$= k^{3} + k^{3} + 1 + 3 k^{2} + 3 k$
$= k^{3} + {(k + 1)}^{3}$

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

The correct option is A p3+(p+1)3a=k2+1Since series is of consecutive integerSum of (2k+1) terms =n2(2a+(n−1)d)=(2k+1)2[2(k2+1)+(2k+1−1)1]=(2k+1)2[2k2+2+2k]=(2k+1)(k2+k+1)=2k3+2k2+2k+k2+k+1=k3+k3+1+3k2+3k=k3+(k+1)3

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