Please solve question 35

Question

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chetan2u · Answer

Hi,
Not the correct way to post a Q..

But still to help you.
The Q is asking the mean of nC0+nC1+....+nCn
The equation above is equal to  2^n  and number of terms from 0,1... to N are n+1

So mean = sum/#= \frac{2^n}{n+1}

Hence A

broall · Answer

Please type the question and format it as the rule of posting.

I made it for you.

What is the answer choice E? This question seems not like GMAT format.

$\text{Mean}=\frac{0 \times C^0_n + 1 \times C^1_n + ... + n \times C^n_n}{C^0_n + C^1_n+ C^2_n+ ... + C^n_n} = \frac{\sum kC^k_n}{\sum C^k_n}$

Note that $C^k_n = \frac{n!}{k!(n-k)!}$ and $C^{n-k}_n = \frac{n!}{(n-k)!k!}$

So $C^k_n=C^{n-k}_n$

Hence we have
$\begin{align}
0C^0_n+nC^n_n &= (0+n)C^0_n &= nC^0_n \
1C^1_n+(n-1)C^{n-1}_n &= (1+n-1)C^{n-1}_n &= nC^1_n \
2C^2_n+(n-2)C^{n-2}_n &= (2+n-2)C^{n-2}_n &= nC^2_n \
... \
kC^k_n+(n-k)C^{n-k}_n &= (k+n-k)C^{n-k}_n &= nC^k_n \
... \
nC^n_n+0C^0_n &= (n+0)C^{n}_n &= nC^n_n \end{align}$
_____________________________________________________
$\implies 2 \sum kC^k_n = \sum nC^k_n $
$ \implies \sum kC^k_n = \frac{n}{2} \times \sum C^k_n$

$\implies \text{Mean} = \frac{\sum kC^k_n}{\sum C^k_n} = \frac{\frac{n}{2} \times \sum C^k_n}{\sum C^k_n}=\frac{n}{2}$

The answer D

Please solve question 35

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