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

Prove that, 3C1+7C2+11C3+.........+(4n−1)Cn=1+(2n−1)⋅2n{}^3{C_1} + {}^7{C_2} + {}^{11}{C_3} + ......... + {}^{(4n - 1)}{C_n} = 1 + \left( {2n - 1} \right) \cdot {2^n}.

Explanation

Solution

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, 3C1+7C2+11C3+.........+(4n−1)Cn=1+(2n−1)⋅2n{}^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
∑r=1n(4r−1)\sum\limits_{r = 1}^n {\left( {4r - 1} \right)} nCr^n{C_r}
As you know
nCr=n!(n−r)!×r!^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)! \times r!}}
⇒∑r=1n(4r−1)\Rightarrow \sum\limits_{r = 1}^n {\left( {4r - 1} \right)} nCr^n{C_r} = ∑r=1n(4r−1)n!(n−r)!×r!\sum\limits_{r = 1}^n {\left( {4r - 1} \right)} \dfrac{{n!}}{{\left( {n - r} \right)! \times r!}}
Now separate the summation
⇒∑r=1n(4r)n!(n−r)!×r!−∑r=1nnCr\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}}
⇒∑r=1n(4r)n(n−1)!(n−r)!×r(r−1)!−(nC1+nC2+nC3+......+nCn)\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 (n−1)!(n−r)!×(r−1)!=n−1Cr−1\dfrac{{\left( {n - 1} \right)!}}{{\left( {n - r} \right)! \times \left( {r - 1} \right)!}}{ = ^{n - 1}}{C_{r - 1}} and (nC0+nC1+nC2+......+nCn−1)=(1+1)n=2n\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 nC0=1^n{C_0} = 1 , So apply this
⇒∑r=1n(4n)\Rightarrow \sum\limits_{r = 1}^n {\left( {4n} \right)} n−1Cr−1−2n^{n - 1}{C_{r - 1}} - {2^n}
⇒4n(n−1C0+n−1C1+n−1C2+......+n−1Cn−1)−(2n−1)\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 (n−1C0+n−1C1+n−1C2+......+n−1Cn−1)=(1+1)n−1=2n−1\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}}
⇒4n×2n−1−2n+1\Rightarrow 4n \times {2^{n - 1}} - {2^n} + 1
⇒2n×2n−2n+1\Rightarrow 2n \times {2^n} - {2^n} + 1
⇒1+(2n−1)2n\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.