Question

Prove that, ${}^3{C_1} + {}^7{C_2} + {}^{11}{C_3} + ......... + {}^{(4n - 1)}{C_n} = 1 + \left( {2n - 1} \right) \cdot {2^n}$.

Answer 1

Hint: Here the above equation is reduced to a series form and then apply a combination formula to prove.

Complete step-by-step answer:
Given, ${}^3{C_1} + {}^7{C_2} + {}^{11}{C_3} + ......... + {}^{(4n - 1)}{C_n} = 1 + \left( {2n - 1} \right) \cdot {2^n}$
Take LHS
This series is written as
$\sum\limits_{r = 1}^n {\left( {4r - 1} \right)} $$^n{C_r}$
As you know
$^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)! \times r!}}$
$ \Rightarrow \sum\limits_{r = 1}^n {\left( {4r - 1} \right)} $$^n{C_r}$ = $\sum\limits_{r = 1}^n {\left( {4r - 1} \right)} \dfrac{{n!}}{{\left( {n - r} \right)! \times r!}}$
Now separate the summation
$ \Rightarrow \sum\limits_{r = 1}^n {\left( {4r} \right)} \dfrac{{n!}}{{\left( {n - r} \right)! \times r!}} - \sum\limits_{r = 1}^n {^n{C_r}} $
$ \Rightarrow \sum\limits_{r = 1}^n {\left( {4r} \right)} \dfrac{{n\left( {n - 1} \right)!}}{{\left( {n - r} \right)! \times r\left( {r - 1} \right)!}} - \left( {^n{C_1}{ + ^n}{C_2}{ + ^n}{C_3} + ......{ + ^n}{C_n}} \right)$
Now you know $\dfrac{{\left( {n - 1} \right)!}}{{\left( {n - r} \right)! \times \left( {r - 1} \right)!}}{ = ^{n - 1}}{C_{r - 1}}$ and $\left( {^n{C_0}{ + ^n}{C_1}{ + ^n}{C_2} + ......{ + ^n}{C_{n - 1}}} \right) = {\left( {1 + 1} \right)^n} = {2^n}$
According to {binomial expansion} and the value of $^n{C_0} = 1$ , So apply this
$ \Rightarrow \sum\limits_{r = 1}^n {\left( {4n} \right)} $$^{n - 1}{C_{r - 1}} - {2^n}$
\[ \Rightarrow 4n\left( {^{n - 1}{C_0}{ + ^{n - 1}}{C_1}{ + ^{n - 1}}{C_2} + ......{ + ^{n - 1}}{C_{n - 1}}} \right) - \left( {{2^n} - 1} \right)\]
You know according to binomial expansion $\left( {^{n - 1}{C_0}{ + ^{n - 1}}{C_1}{ + ^{n - 1}}{C_2} + ......{ + ^{n - 1}}{C_{n - 1}}} \right) = {\left( {1 + 1} \right)^{n - 1}} = {2^{n - 1}}$
\[ \Rightarrow 4n \times {2^{n - 1}} - {2^n} + 1\]
\[ \Rightarrow 2n \times {2^n} - {2^n} + 1\]
\[ \Rightarrow 1 + \left( {2n - 1} \right){2^n}\] = RHS
Hence proved.

Note: In this type of question answer can be in any form but take care what you have to prove, you have to give an answer in that form.