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Given that ${{a}_{1}},{{a}_{2}},{{a}_{3}},......{{a}_{n}}$ form an arithmetic progression, find the following sum $S=\sum\limits_{i=1}^{n}{\dfrac{{{a}_{i}}{{a}_{i+1}}{{a}_{i+2}}}{{{a}_{i}}+{{a}_{i+2}}}}$
(A) $S=\dfrac{n}{2}\left[ a_{1}^{2}+{{a}_{1}}d\left( n \right)+\dfrac{\left( n-2 \right)\left( 2n+5 \right)}{6}{{d}^{2}} \right]$
(B) $S=\dfrac{n}{2}\left[ a_{1}^{2}+{{a}_{1}}d\left( n+1 \right)+\dfrac{\left( n-2 \right)\left( 2n+5 \right)}{6}{{d}^{2}} \right]$
(C) $S=\dfrac{n}{2}\left[ a_{1}^{2}+{{a}_{1}}d\left( n+1 \right)+\dfrac{\left( n-1 \right)\left( 2n+5 \right)}{6}{{d}^{2}} \right]$
(D) $S=\dfrac{n}{2}\left[ a_{1}^{3}+{{a}_{1}}d\left( n+1 \right)+\dfrac{\left( n-1 \right)\left( 2n+5 \right)}{6}{{d}^{2}} \right]$

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Last updated date: 20th Jun 2024
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Answer
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Hint: For answering this question we will use the basic concept regarding the arithmetic progressions. When given ${{a}_{1}},{{a}_{2}},.............{{a}_{n}}$ forms arithmetic progression we can say that ${{a}_{i+1}}={{a}_{i}}+d$ and ${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$ where $d={{a}_{2}}-{{a}_{1}}$ and the sum of the $n$ term which is given as ${{S}_{n}}=\sum\limits_{i=1}^{n}{{{a}_{i}}}= \left( \dfrac{n}{2} \right)\left[ 2{{a}_{1}}+\left( n -1 \right)d \right]$

Complete step-by-step solution:
Now considering from the question we have been given that ${{a}_{1}},{{a}_{2}},.............{{a}_{n}}$ forms arithmetic progression. So we can say that ${{a}_{i+1}}={{a}_{i}}+d$ because ${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$ where $d={{a}_{2}}-{{a}_{1}}$ .
For answering this question we need to find the sum $S=\sum\limits_{i=1}^{n}{\dfrac{{{a}_{i}}{{a}_{i+1}}{{a}_{i+2}}}{{{a}_{i}}+{{a}_{i+2}}}}$
By substituting the respective values we will have,
$\begin{align}
  & S=\sum\limits_{i=1}^{n}{\dfrac{{{a}_{i}}\left( {{a}_{i}}+d \right)\left( {{a}_{i}}+2d \right)}{{{a}_{i}}+{{a}_{i}}+2d}} \\
 & \Rightarrow S=\sum\limits_{i=1}^{n}{\dfrac{{{a}_{i}}\left( {{a}_{i}}+d \right)\left( {{a}_{i}}+2d \right)}{2\left( {{a}_{i}}+d \right)}} \\
 & \Rightarrow S=\dfrac{1}{2}\left[ \sum\limits_{i=1}^{n}{a_{i}^{2}+\sum\limits_{i=1}^{n}{2{{a}_{i}}d}} \right] \\
\end{align}$
Now, we will write the expansion of $\sum\limits_{i=1}^{n}{a_{i}^{2}}$
$\begin{align}
  & \sum\limits_{i=1}^{n}{a_{i}^{2}}={{\sum\limits_{i=1}^{n}{\left( {{a}_{i}}+\left( i-1 \right)d \right)}}^{2}} \\
 & \Rightarrow \sum\limits_{i=1}^{n}{\left( a_{i}^{2}+{{d}^{2}}{{\left( i-1 \right)}^{2}}+2{{a}_{i}}d\left( i-1 \right) \right)} \\
\end{align}$
By expanding the terms $\sum\limits_{i=1}^{n}{\left( i-1 \right)}$ and $\sum\limits_{i=1}^{n}{{{\left( i-1 \right)}^{2}}}$, we will get the below equation using $\sum\limits_{i=1}^{n}{i}=\dfrac{n\left( n-1 \right)}{2}$ and $\sum\limits_{i=1}^{n}{{{i}^{2}}}=\dfrac{n\left( n-1 \right)\left( 2n-1 \right)}{6}$ we will have $\sum\limits_{i=1}^{n}{a_{i}^{2}}=na_{i}^{2}+{{d}^{2}}\left( \dfrac{n\left( n-1 \right)\left( 2n-1 \right)}{6} \right)+2{{a}_{i}}d\left( \dfrac{n\left( n-1 \right)}{2} \right)$
We will also use the formulae for the sum of $n$ terms which is given as ${{S}_{n}}=\sum\limits_{i=1}^{n}{{{a}_{i}}}= \left( \dfrac{n}{2} \right)\left[ 2{{a}_{1}}+\left( n -1 \right)d \right] $
Now, by using this value, we will get the below equation. After substituting we will simplify the equation more further very carefully.
$\begin{align}
  & S=\dfrac{1}{2}\left[ na_{1}^{2}+{{d}^{2}}\left( \dfrac{n\left( n-1 \right)\left( 2n-1 \right)}{6} \right)+2{{a}_{1}}d\left( \dfrac{n\left( n-1 \right)}{2} \right) \right]+2d\left( \dfrac{n}{2} \right)\left[ 2{{a}_{1}}+\left( n -1 \right)d \right] \\
 & \Rightarrow S=\dfrac{1}{2}\left[ na_{1}^{2}+{{d}^{2}}\left( \dfrac{n\left( n -1 \right)\left( 2n-1 \right)}{6} \right)+2{{a}_{1}}d\left( \dfrac{n\left( n-1 \right)}{2} \right) \right]+2{{a}_{1}}dn+{{d}^{2}}n\left( n -1 \right) \\
 & \Rightarrow S=\dfrac{1}{2}\left[ na_{1}^{2}+{{d}^{2}}n\left( n-1 \right)\left[ \dfrac{2n-1}{6}+1 \right]+2{{a}_{1}}dn\left[ \dfrac{n -1}{2}+1 \right] \right] \\
 & \Rightarrow S=\dfrac{1}{2}\left[ na_{1}^{2}+\left( \dfrac{{{d}^{2}}n\left( n-1 \right)\left( 2n+5 \right)}{6} \right)+{{a}_{1}}dn\left( n+1 \right) \right] \\
 & \Rightarrow S=\dfrac{n}{2}\left[ a_{1}^{2}+\left( \dfrac{{{d}^{2}}\left( n -1 \right)\left( 2n+5 \right)}{6} \right)+{{a}_{1}}d\left( n+1 \right) \right] \\
\end{align}$
Therefore option C is the correct option.

Note: While answering questions of this type we should be sure with the concept and expansions we make. Similar to arithmetic progression there exist geometric progression when given ${{a}_{1}},{{a}_{2}},.............{{a}_{n}}$ forms geometric progression we can say that ${{a}_{i+1}}={{a}_{i}}r$ and ${{a}_{n}}={{a}_{1}}{{r}^{n-1}}$ where $r=\dfrac{{{a}_{2}}}{{{a}_{1}}}$ and the sum of the $n$ term which is given as ${{S}_{n}}=\sum\limits_{i=1}^{n}{{{a}_{i}}}=\dfrac{{{a}_{1}}\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}$.