\[
\left[\begin{array}{l}
\text { Pascal's Triangle and the Binomial Theorem } \\
{[\underline{\text { Induction (other) }}]}
\end{array}\right.
\]

\[
\begin{aligned}
& \frac{1}{1} \\
& \frac{1}{2} \quad \frac{1}{2} \\
& \frac{1}{3} \quad \frac{1}{6} \quad \frac{1}{3} \\
& \begin{array}{llll}
\frac{1}{4} & \frac{1}{12} & \frac{1}{12} & \frac{1}{4}
\end{array} \\
& \begin{array}{lllll}
\frac{1}{5} & \frac{1}{20} & \frac{1}{30} & \frac{1}{20} & \frac{1}{5}
\end{array} \\
& \begin{array}{llllll}
\frac{1}{6} & \frac{1}{30} \quad \frac{1}{60} \quad \frac{1}{60} \quad \frac{1}{30} \quad \frac{1}{6}
\end{array}
\end{aligned}
\]

Here is a fragment of a table known as Leibniz's Triangle. Its properties are "analogous in the sense of opposition" to those of Pascal's Triangle. The numbers on the boundary of the triangle are the reciprocals of consecutive natural numbers. Each number inside is equal to the sum of the two numbers below it. Find the formula that connects the numbers from Pascal's Triangle and Leibniz's Triangle.
The denominators of the numbers located in the rows of the harmonic triangle are proportional to the elements of Pascal's triangle, with the boundary members serving as the coefficients of proportionality. Where Pascal's triangle has the number $C_{n}^{k}$, Leibniz's triangle contains $\frac{1}{(n+1) C_{n}^{k}}$. The recurrence formula $\frac{1}{(n+1) C_{n}^{k-1}}+\frac{1}{(n+1) C_{n}^{k}}=\frac{1}{n C_{n-1}^{k-1}}$ can be verified by direct calculation.

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