Question

If \[\dfrac{{}^{n}{{C}_{0}}}{{{2}^{n}}}+2.\dfrac{{}^{n}{{C}_{1}}}{{{2}^{n}}}+3.\dfrac{{}^{n}{{C}_{2}}}{{{2}^{n}}}+......+\left( n+1 \right)\dfrac{{}^{n}{{C}_{n}}}{{{n}^{n}}}=16\], then the value of ‘n’ is: -
(a) 20
(b) 25
(c) 30
(d) 40

Answer 1

Hint: Take \[{{2}^{n}}\] common from all term. Inside the bracket, write, \[{}^{n}{{C}_{0}}+2.{}^{n}{{C}_{1}}+3.{}^{n}{{C}_{2}}+....+\left( n+1 \right){}^{n}{{C}_{n}}\] in the form \[{}^{n}{{C}_{0}}+\left( 1+1 \right).{}^{n}{{C}_{1}}+\left( 1+2 \right).{}^{n}{{C}_{2}}+....+\left( 1+n \right){}^{n}{{C}_{n}}\]. Multiply all the terms and group \[\left( {}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+....+{}^{n}{{C}_{n}} \right)\] together whose value is \[{{2}^{n}}\]. Now, the terms left will be \[\left( {}^{n}{{C}_{0}}+2.{}^{n}{{C}_{1}}+3.{}^{n}{{C}_{2}}+....+n.{}^{n}{{C}_{n}} \right)\] whose value we have to find. Assume this expression as S. Again write S by adding the terms together. From here find the value of S and put in the expression to solve for the value of n.

Complete step by step answer:
We have been given: - \[\dfrac{{}^{n}{{C}_{0}}}{{{2}^{n}}}+2.\dfrac{{}^{n}{{C}_{1}}}{{{2}^{n}}}+3.\dfrac{{}^{n}{{C}_{2}}}{{{2}^{n}}}+......+\left( n+1 \right)\dfrac{{}^{n}{{C}_{n}}}{{{n}^{n}}}=16\].
This can be written as: -
\[\begin{align}
  & \Rightarrow \dfrac{1}{{{2}^{n}}}\left[ {}^{n}{{C}_{0}}+\left( 1+1 \right){}^{n}{{C}_{1}}+\left( 1+2 \right){}^{n}{{C}_{2}}+....+\left( 1+n \right){}^{n}{{C}_{n}} \right]=16 \\
 & \Rightarrow \dfrac{1}{{{2}^{n}}}\left[ \left( {}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+{}^{n}{{C}_{n}} \right) + \left( 1.{}^{n}{{C}_{1}}+2.{}^{n}{{C}_{2}}+.....+n.{}^{n}{{C}_{n}} \right) \right]=16 \\
\end{align}\]
We know that, \[{}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+.....+{}^{n}{{C}_{n}}={{2}^{n}}\], therefore,
\[\Rightarrow \dfrac{1}{{{2}^{n}}}\left[ {{2}^{n}}+1.{}^{n}{{C}_{1}}+2.{}^{n}{{C}_{2}}+.....+n.{}^{n}{{C}_{n}} \right]=16\]
Let us assume: -
\[\Rightarrow S=1.{}^{n}{{C}_{1}}+2.{}^{n}{{C}_{2}}+.....+n.{}^{n}{{C}_{n}}\]
\[\Rightarrow S=0.{}^{n}{{C}_{0}}+1.{}^{n}{{C}_{1}}+2.{}^{n}{{C}_{2}}+.....+n.{}^{n}{{C}_{n}}\] - (i)
This can also be written as: -
\[\Rightarrow S=n.{}^{n}{{C}_{n}}+\left( n-1 \right).{}^{n}{{C}_{n-1}}+........+1.{}^{n}{{C}_{1}}+0.{}^{n}{{C}_{0}}\] - (ii)
Now, adding equation (i) and (ii), we get,
\[\Rightarrow 2S=\left( 0.{}^{n}{{C}_{0}}+n.{}^{n}{{C}_{n}} \right)+\left( 1.{}^{n}{{C}_{1}}+\left( n-1 \right).{}^{n}{{C}_{n-1}} \right)+\left( 2.{}^{n}{{C}_{2}}+\left( n-2 \right).{}^{n}{{C}_{n-2}} \right)+.....+\left( n.{}^{n}{{C}_{n}}+0.{}^{n}{{C}_{0}} \right)\]
Using the equality, \[{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}\], we get,
\[\begin{align}
  & \Rightarrow 2S=\left( 0.{}^{n}{{C}_{0}}+n.{}^{n}{{C}_{n}} \right)+\left( 1.{}^{n}{{C}_{1}}+\left( n-1 \right).{}^{n}{{C}_{1}} \right)+\left( 2.{}^{n}{{C}_{2}}+\left( n-2 \right).{}^{n}{{C}_{2}} \right)+.....+\left( n.{}^{n}{{C}_{n}}+0.{}^{n}{{C}_{0}} \right) \\
 & \Rightarrow 2S=n.{}^{n}{{C}_{0}}+n.{}^{n}{{C}_{1}}+n.{}^{n}{{C}_{2}}+......+n.{}^{n}{{C}_{n}} \\
 & \Rightarrow 2S=n\left[ {}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+......+{}^{n}{{C}_{n}} \right] \\
 & \Rightarrow 2S=n{{.2}^{n}} \\
 & \Rightarrow S=n{{.2}^{n-1}} \\
\end{align}\]
Substituting the value of S in the required expression, we get,
\[\begin{align}
  & \Rightarrow \dfrac{1}{{{2}^{n}}}\left[ {{2}^{n}}+n{{.2}^{n-1}} \right]=16 \\
 & \Rightarrow 1+\dfrac{n}{2}=16 \\
 & \Rightarrow \dfrac{n}{2}=15 \\
 & \Rightarrow n=30 \\
\end{align}\]

So, the correct answer is “Option C”.

Note: One may note that we can also remember the result of ‘S’, that will help us to solve these types of questions easily and in less time. You may see that we have added \[0.{}^{n}{{C}_{0}}\] in the expression of S, this is because we have to group the terms so that a pattern can be formed. The term \[0.{}^{n}{{C}_{0}}\] has the value 0 and therefore it does not affect the value of S.