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)
Integer Answer Type Question
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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 $s _n=1+q+q^{2}+\ldots+q^{n}=\frac{1-q^{n+1}}{1-q}$
$$ \begin{aligned} \therefore \sum _{r=1}^{n+1}{ }^{n+1} C _r \frac{1-q^{r}}{1-q} & =\frac{1}{1-q} \sum _{r=1}^{n+1}{ }^{n+1} C _r-\sum _{r=1}^{n+1}{ }^{n+1} C _r q^{r} \\ & =\frac{1}{1-q}\left[(1+1)^{n+1}-(1+q)^{n+1}\right] \\ & =\frac{1}{1-q}\left[2^{n+1}-(1+q)^{n+1}\right] . \end{aligned} $$
$$ =\frac{1-\frac{q+1}{2}^{n+1}}{1-\frac{q+1}{2}}=\frac{2^{n+1}-(q+1)^{n+1}}{2^{n}(1-q)} $$
From Eqs. (i) and (ii),
$$ { }^{n+1} C _1+{ }^{n+1} C _2 s _1+{ }^{n+1} C _3 s _2+\ldots+{ }^{n+1} C _{n+1} s _n=2{ }^{n} S _n $$
