Binomial Theorem 2 Question 11
14. For any positive integers $m, n$ (with $n \geq m$ ),
If ${ } _m^{n}={ }^{n} C _m$. Prove that
$$ \frac{n}{m}+\begin{gathered} n-1 \\ m \end{gathered}+\begin{gathered} n-2 \\ m \end{gathered}+\ldots+{ } _m^{m}=\begin{aligned} & n+1 \\ & m+1 \end{aligned} $$
Prove that
or
$$ \begin{aligned} & \begin{array}{c} n \\ m \end{array}+2 \begin{array}{c} n-1 \\ m \end{array}+3{ } _m^{n-2}+\ldots+(n-m+1) \\ & m=n+2 \end{aligned} $$
(IIT JEE 2000, 6M)
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Solution:
- Let $S={ } _m^{n}+{ } _m^{n-1}+{ } _m^{n-2}+\ldots+{ } _m^{m}=\begin{aligned} & n+1 \ & m+1\end{aligned}$
It is obvious that, $n \geq m$.
[given]
NOTE This question is based upon additive loop.
Now, $S={ } _m^{m}+{ } _m^{m+1}+{ } _m^{m+2}+\ldots+{ } _m^{n}$
$$ \begin{aligned} & =\begin{array}{c} m+1 \\ m+1 \end{array}+\begin{array}{c} m+1 \\ m \end{array}+\begin{array}{c} m+2 \\ m \end{array}+\ldots \begin{array}{c} n \\ m \end{array} \\ & \because{ } _m^{m}=1=\begin{array}{l} m+1 \\ m+1 \end{array} \\ & =\frac{m+2}{m+1}+\frac{m+2}{m}+\ldots+\begin{array}{c} n \\ m \end{array} \\ & ={ } _{m+1}^{m+3}+\ldots+\frac{n}{m} \\ & =\ldots \ldots \ldots \ldots \ldots \ldots . . . . . . . . . . . . . . . . \\ & ={ } _{m+1}^{n}+{ } _m^{n}=\begin{array}{c} n+1 \\ m+1 \end{array} \text {, which is true. } \end{aligned} $$
Again, we have to prove that
$$ \begin{aligned} & \begin{array}{c} n \\ m \end{array}+2 \begin{array}{c} n-1 \\ m \end{array}+3 \begin{array}{c} n-2 \\ m \end{array}+\ldots+(n-m+1) \quad \begin{array}{l} m \\ m \end{array}=\begin{array}{l} n+2 \\ m+2 \end{array} \\ & \text { Let } S _1={ } _m^{n}+2 \begin{array}{cc} n-1 \\ m \end{array}+3 \begin{array}{c} n-2 \\ m \end{array}+\ldots+(n-m+1){ } _m^{m} \\ & =\frac{n}{m}+\frac{n-1}{m}+\frac{n-2}{m}+\ldots+\frac{m}{m} \\ & +\begin{array}{c} n-1 \\ m \end{array}+\frac{n-2}{m}+\ldots+\begin{array}{l} m \\ m \end{array} \\ & +\begin{array}{c} n-2 \\ m \end{array}+\ldots+{ } _m^{m} n-m+1 \text { rows } \\ & \begin{array}{c} \cdots \cdots \\ +\quad \begin{array}{c} m \\ m \end{array} \end{array} \end{aligned} $$
Now, sum of the first row is $n+1$
$m+1$
Sum of the second row is $\begin{gathered}n \ m+1\end{gathered}$.
Sum of the third row is $n-1$
$m+1$,
Sum of the last row is $\begin{aligned} & m \ & m\end{aligned}=\begin{aligned} & m+1 \ & m+1\end{aligned}$.
Thus, $\quad S=\begin{gathered}n+1 \ m+1\end{gathered}+\begin{gathered}n \ m+1\end{gathered}+\begin{gathered}n-1 \ m+1\end{gathered}$
$$ +\ldots+\begin{gathered} m+1 \\ m+1 \end{gathered}=\begin{gathered} n+1+1 \\ m+2 \end{gathered}=\begin{gathered} n+2 \\ m+2 \end{gathered} $$
[from Eq. (i) replacing $n$ by $n+1$ and $m$ by $m+1$ ]
