Sum to n Terms of Special Series - Practice Questions & MCQ

The sum of some special series ( $V_n$ method)
The sum of the series of the form

$
\begin{aligned}
& a_1 a_2 \ldots \ldots a_r+a_2 a_3 \ldots \ldots a_{r+1}+\ldots \ldots \ldots+a_n a_{n+1} \ldots \ldots a_{n+r-1} \\
& S_n=a_1 a_2 \ldots \ldots a_r+a_2 a_3 \ldots \ldots a_{r+1}+\ldots \ldots \ldots \ldots+a_n a_{n+1} \ldots \ldots a_{n+r-1} \\
& T_n=a_n a_{n+1} \ldots \ldots \ldots \ldots a_{n+r-2} a_{n+r-1}
\end{aligned}
$
Let $V_n=a_n a_{n+1} \ldots \ldots \ldots \ldots a_{n+r-2} a_{n+r-1} a_{n+r} \quad$ [Taking one extra factor in $T_n$ for $V_n$ ]

$
\begin{aligned}
& V_{n-1}=a_{n-1} a_n a_{n+1} \ldots \ldots \ldots a_{n+r-3} a_{n+r-2} a_{n+r-1} \\
& \Rightarrow V_n-V_{n-1}=a_n a_{n+1} a_{n+2} \ldots \ldots a_{n+r-1}\left(a_{n+r}-a_{n-1}\right)=T_n\left(a_{n+r}-a_{n-1}\right)
\end{aligned}
$

$d$ be the common difference of an AP, then

$
\begin{aligned}
& a_n=a_1+(n-1) d \\
& V_n-V_{n-1}=T_n\left[\left\{a_1+(n+r-1) d\right\}-\left\{a_1+(n-2) d\right\}\right]=(r+1) d \cdot T_n \\
& \Rightarrow T_n=\frac{1}{(r+1) d}\left(V_n-V_{n-1}\right) \\
& S_n=\sum T_n=\frac{1}{(r+1) d} \sum_{n=1}^n\left(V_n-V_{n-1}\right)
\end{aligned}
$

$S_n=\frac{1}{(r+1) d}\left(V_n-V_0\right) \quad$ [from the telescoping series]

$
S_n=\frac{1}{(r+1)\left(a_2-a_1\right)}\left(a_n a_{n+1} a_{n+2} \ldots a_{n+r}-a_0 a_1 a_2 \ldots \ldots a_r\right)
$
If $a_1, a_2, a_3, \ldots \ldots \ldots, a_n$ are in AP. Then

$
\begin{aligned}
& \bullet a_1 a_2+a_2 a_3+\ldots \ldots \ldots+a_n a_{n+1}=\frac{1}{3\left(a_2-a_1\right)}\left(a_n a_{n+1} a_{n+2}-a_0 a_1 a_2\right) \\
& \bullet a_1 a_2 a_3+a_2 a_3 a_4+\ldots \ldots \ldots+a_n a_{n+1} a_{n+2}=\frac{1}{4\left(a_2-a_1\right)}\left(a_n a_{n+1} a_{n+2} a_{n+2}-a_0 a_1 a_2 a_4\right)
\end{aligned}
$
 

The sum of some special series ( $V_n$ method)
The sum of the series of the form

$
\begin{aligned}
& \frac{1}{a_1 a_2 \ldots a_r}+\frac{1}{a_2 a_3 \ldots a_{r+1}}+\ldots \ldots+\frac{1}{a_n a_{n+1} \ldots a_{n+r-1}} \\
& S_n=\frac{1}{a_1 a_2 \ldots \ldots a_r}+\frac{1}{a_2 a_3 \ldots a_{r+1}}+\ldots \ldots+\frac{1}{a_n a_{n+1 \ldots \ldots a_{n+r-1}}} \\
& T_n=\frac{1}{a_n a_{n+1} \ldots \ldots a_{n+r-2} a_{n+r-1}}
\end{aligned}
$
Let, $V_n=\frac{1}{a_{n+1} a_{n+2} \ldots \ldots a_{n+r-2} a_{n+r-1}}$ [Leaving first factor from the denominator of $T_n$ ]
So, $V_{n-1}=\frac{1}{a_n a_{n+1} \ldots \ldots . . . a_{n+r-3} a_{n+r-2}}$

$
\begin{aligned}
& \Rightarrow V_n-V_{n-1}=\frac{1}{a_{n+1} a_{n+2} \ldots \ldots a_{n+r-2} a_{n+r-1}}-\frac{1}{a_n a_{n+1} \ldots \ldots a_{n+r-3} a_{n+r-2}} \\
& T_n=\frac{\left(V_n-V_{n-1}\right)}{d(1-r)} \\
& S_n=\sum T_n=\sum_{n=1}^n \frac{\left(V_n-V_{n-1}\right)}{d(1-r)}=\frac{1}{d(1-r)}\left(V_n-V_0\right) \\
& S_n=\frac{1}{(r-1)\left(a_2-a_1\right)}\left\{\frac{1}{a_1 a_2 \ldots a_{r-1}}-\frac{1}{a_{n+1} a_{n+2 \ldots a_{n+r-1}}}\right\}
\end{aligned}
$
If $a_1, a_2, a_3, \ldots \ldots \ldots, a_n$ are in AP. Then

$
\begin{aligned}
& \bullet \frac{1}{a_1 a_2}+\frac{1}{a_2 a_3}+\ldots \ldots+\frac{1}{a_n a_{n+1}}=\frac{n}{a_1 a_{n+1}} \\
& \bullet \frac{1}{a_1 a_2 a_3}+\frac{1}{a_2 a_3 a_4}+\ldots \ldots+\frac{1}{a_n a_{n+1} a_{n+2}}=\frac{1}{2\left(a_2-a_1\right)}\left\{\frac{1}{a_1 a_2}-\frac{1}{a_{n+1} a_{n+2}}\right\}
\end{aligned}
$