1/(11!+12!)+1/(12!+13!)+1/(13!+14!)+⋯+1/(31!+32!)

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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