Binomial Theorem 1 Question 33
35.
Given, $s_{n}=1+q+q^{2}+\ldots+q^{n}$
Prove that ${ }^{n+1} C_{1}+{ }^{n+1} C_{2} s_{1}+{ }^{n+1} C_{3} s_{2}$ $ +\ldots+{ }^{n+1} C_{n+1} s_{n}=2^{n} S_{n} $
$(1984,4$ M)
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Solution:
- ${ }^{n+1} C_{1}+{ }^{n+1} C_{2} s_{1}+{ }^{n+1} C_{3} s_{2}+\ldots+{ }^{n+1} C_{n+1} s_{n}$
$ =\sum_{r=1}^{n+1}{ }^{n+1} C_{r} s_{r-1} $
where $s_{n}=1+q+q^{2}+\ldots+q^{n}=\frac{1-q^{n+1}}{1-q}$
$ \begin{aligned} \therefore \sum_{r=1}^{n+1}{ }^{n+1} C_{r} \frac{1-q^{r}}{1-q} & =\frac{1}{1-q} \sum_{r=1}^{n+1}{ }^{n+1} C_{r}-\sum_{r=1}^{n+1}{ }^{n+1} C_{r} q^{r} \\ & =\frac{1}{1-q}\left[(1+1)^{n+1}-(1+q)^{n+1}\right] \\ & =\frac{1}{1-q}\left[2^{n+1}-(1+q)^{n+1}\right] . \quad …….(i) \end{aligned} $
$ =\frac{1-(\frac{q+1}{2})^{n+1}}{1-(\frac{q+1}{2})}=\frac{2^{n+1}-(q+1)^{n+1}}{2^{n}(1-q)} \quad …….(ii) $
From Eqs. (i) and (ii),
$ { }^{n+1} C_{1}+{ }^{n+1} C_{2} s_{1}+{ }^{n+1} C_{3} s_{2}+\ldots+{ }^{n+1} C_{n+1} s_{n}=2{ }^{n} S_{n} $
