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Binomial Theorem Part 1

Section: Mathematics

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Question : 93 of 100

Marks: +1, -0

If

S_n=\sum\limits_{r=0}^{n} \;\frac{1}{{}^{n}C_{r}}

t_n=\sum\limits_{r=0}^{n} \;\frac{r}{{}^{n}C_{r}}

\;\frac{t_n}{S_n}

$\frac{2n-1}{2}$
$\frac{1}{2}n-1$
$n-1$
$\frac{1}{2}n$

Solution:

S_n=\sum\limits_{r=0}^{n} \;\frac{1}{{}^{n}C_{r}}

=\;\frac{1}{{}^{n}C_{0}}+\;\frac{1}{{}^{n}C_{1}}+ \dots +\frac{1}{{}^{n}C_{n}}

t_n=\sum\limits_{r=0}^{n} \;\frac{r}{{}^{n}C_{r}}

=\;\frac{0}{{}^{n}C_{0}}+\;\frac{1}{{}^{n}C_{1}}+\;\frac{2}{{}^{n}C_{2}} \dots +\frac{n}{{}^{n}C_{n}}

......(1)

We can write

t_n

by rearranging like this,

t_n=\;\frac{n}{{}^{n}C_{n}}+\;\frac{n-1}{{}^{n}C_{n-1}}+ \dots +\frac{1}{{}^{n}C_{1}}+\frac{0}{{}^{n}C_{0}}

=\;\frac{n}{{}^{n}C_{0}}+\;\frac{n-1}{{}^{n}C_{1}}+ \dots +\frac{1}{{}^{n}C_{n-1}}+\frac{0}{{}^{n}C_{n}}

......(2)

[as

{}^{n}C_{0}={}^{n}C_{n},{}^{n}C_{1}={}^{n}C_{n-1} \dots ]

By adding (1) and (2) we get,

\;2 t_n=\;\frac{n}{{}^{n}C_{0}}+\;\frac{n}{{}^{n}C_{1}}+ \dots +\frac{n}{{}^{n}C_{n-1}}+\frac{n}{{}^{n}C_{n}}

\;= n \left[ \frac{1}{{}^{n}C_{0}}+\frac{1}{{}^{n}C_{1}}+ \dots +\frac{1}{{}^{n}C_{n-1}}+\frac{1}{{}^{n}C_{n}} \right]

\;= n S_n

\; \therefore \; \frac{t_n}{S_n} = \frac{n}{2}

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