The first term of an A.P. of consecutive inte | vedclass

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(D) Given that the $A.P.$ consists of consecutive integers,the common difference $d = 1$.The first term $a = {p^2} + 1$.The number of terms $n = 2p +

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

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(D) Given that the $A.P.$ consists of consecutive integers,the common difference $d = 1$.The first term $a = {p^2} + 1$.The number of terms $n = 2p + 1$.The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2} \{ 2a + (n - 1)d \}$.Substituting the values:$S_{2p+1} = \frac{2p+1}{2} \{ 2({p^2} + 1) + (2p + 1 - 1)(1) \}$$S_{2p+1} = \frac{2p+1}{2} \{ 2{p^2} + 2 + 2p \}$$S_{2p+1} = (2p + 1)({p^2} + p + 1)$Expanding the expression:$(2p + 1)({p^2} + p + 1) = 2{p^3} + 2{p^2} + 2p + {p^2} + p + 1 = 2{p^3} + 3{p^2} + 3p + 1$Now,consider the expression ${p^3} + {(p + 1)^3}$:${p^3} + ({p^3} + 3{p^2} + 3p + 1) = 2{p^3} + 3{p^2} + 3p + 1$.Thus,the sum is ${p^3} + {(p + 1)^3}$.

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

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