Question

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

• A. ${(p+1)}^2$
• B. ${(p+1)}^3$
• C. $(2p+1){(p+1)}^2$
• D. $p^3+{(p+1)}^3$

Answer 1

The correct option is D

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

Calculate the sum of the given Arithmetic progression

Given: First term of A.P, $a=p^2+1$

Here $d$ (common difference) $=1$, since integers are consecutive.

Sum of first $n$ terms of A.P, $S_n=\frac{\mathrm{n}}{2}(2\mathrm{a}+\hspace{0.17em}(\mathrm{n}-1\left)\mathrm{d}\right)$

$S_{2p+1}=\left(\frac{2\mathrm{p}+1}{2}\right)\left(2\mathrm{a}+\hspace{0.17em}(2\mathrm{p}+\hspace{0.17em}1-1)\mathrm{d}\right)$

$$\begin{gathered} \text{=}\left(\frac{2\mathrm{p}+1}{2}\right)(2({\mathrm{p}}^2+\hspace{0.17em}1)+2\mathrm{p}) \\ =(2\mathrm{p}\hspace{0.17em}+1)({\mathrm{p}}^2+\hspace{0.17em}1+\mathrm{p}) \\ =2{\mathrm{p}}^3+{\mathrm{p}}^2+\hspace{0.17em}2\mathrm{p}+1+2{\mathrm{p}}^2+\mathrm{p} \\ ={\mathrm{p}}^3+{\mathrm{p}}^3+3{\mathrm{p}}^2\hspace{0.17em}+3\mathrm{p}+\hspace{0.17em}1 \\ ={\mathrm{p}}^3+{\left(\mathrm{p}\hspace{0.17em}+1\right)}^3[\because {(a+b)}^3=a^3+b^3+3a^{2}b+3b^{2}a] \end{gathered}$$

Hence, the sum of$(2p+1)$ terms of the given series can be expressed as $p^3+{(p+1)}^3$, so option (D) is the correct answer.