What is the sum of the following sequence?
\displaystyle {\frac{1}{11!+12!} + \frac{1}{12!+13!} +\frac{1}{13!+14!} +\cdots + \frac{1}{31!+32!}}
a) \displaystyle {\frac{1}{11!} - \frac{1}{32!} }
b) \displaystyle {\frac{1}{12!} - \frac{1}{32!} }
c) \displaystyle {\frac{1}{10!} - \frac{1}{32!} }
d) \displaystyle {\frac{1}{12!} - \frac{1}{33!} }
Ans: d) \displaystyle {\frac{1}{12!} - \frac{1}{33!} }
Explanation:
Concept of n! is required to solve this type of problem.
n! =n\times (n-1)!=n\times (n-1)\times (n-2)! etc.
\begin{align*}
\frac{1}{11!+12!}&=\frac{1}{11!+12\times 11!}\\
&=\frac{1}{(1+12)\times 11!}\\
&=\frac{1}{13\times 11!}\\
&=\frac{12}{13\times 12\times 11!}\\
&=\frac{12}{13!}\\
&=\frac{13-1}{13!}\\
&=\frac{13}{13!}-\frac{1}{13!}\\
&=\frac{13}{13\times 12!}-\frac{1}{13!}\\
&=\frac{1}{12!}-\frac{1}{13!}\\
\end{align*}
\begin{align*}
\frac{1}{12!+13!}&=\frac{1}{12!+13\times 12!}\\
&=\frac{1}{(1+13)\times 12!}\\
&=\frac{1}{14\times 12!}\\
&=\frac{13}{14\times 13\times 12!}\\
&=\frac{13}{14!}\\
&=\frac{14-1}{14!}\\
&=\frac{1}{13!}-\frac{1}{14!}\\
\end{align*}
Similarly, \begin{align*} \frac{1}{13!+14!} &=\frac{1}{14!}-\frac{1}{15!}\\ \end{align*}
And, \begin{align*} \frac{1}{31!+32!} &=\frac{1}{32!}-\frac{1}{33!}\\ \end{align*}
Now writing the full expression:
\begin{align*}
&\frac{1}{11!+12!} + \frac{1}{12!+13!} +\frac{1}{13!+14!} +\cdots + \frac{1}{31!+32!}\\
&=\frac{1}{12!}-\frac{1}{13!} + \frac{1}{13!}-\frac{1}{14!} + \frac{1}{14!}-\frac{1}{15!} + \cdots + \frac{1}{32!}-\frac{1}{33!}\\
&=\frac{1}{12!}-\frac{1}{33!}
\end{align*}
Except first and last terms, all will be cancelled out.
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