Animated Solution for Mathematics - Sequence and Series: If $\frac{1}{2\times 3\times 4}+\frac{1}{3\times 4\times 5}+\frac{1}{4\times 5\times 6}+\cdots +\frac{1}{100\times 101\times 102}=\frac{k}{101}$, then $34k$ is equal to ________.

The given series is:

This is a classic telescoping series. The general term $T_n$ for this series, where $n$ ranges from $1$ to $99$, is defined as:

To solve this, we employ the Method of Differences (or the $V_n$ method). We aim to express $T_n$ as the difference of two consecutive terms, $V_n-V_{n+1}$.

Observe the outer factors of the denominator: $(n+3)$ and $(n+1)$. Their difference is $(n+3)-(n+1)=2$.

We multiply and divide the expression by $2$ to introduce this difference into the numerator:

Splitting this into two separate fractions reveals the telescoping structure:

Simplifying the expression, we obtain:

As we sum these terms from $n=1$ to $n=99$, the intermediate terms cancel out. We are left only with the first positive fraction and the last negative fraction:

We simplify the sum $S$:

Since $102=6\times 17$, we find a common denominator:

Dividing the numerator and denominator by $6$, we get:

Given the form $S=\frac{k}{101}$, we identify $k=\frac{143}{17}$. Therefore, the final value is: