Question

If C denotes the binomial coefficient $^{n}{{C}_{r}}$then \left( -1 \right)C_{0}^{2}+2C_{1}^{2}+5C_{2}^{2}+...+\left( 3n-1 \right)C_{n}^{2}=$$$$$ A.\left( 3n-2 \right)\left( ^{2n}{{C}{n}} \right) B. $\left( \dfrac{3n-2}{2} \right)\left( ^{2n}{{C}_{n}} \right)
C. {{\left( 5+3n \right)}^{2n}}\left( ^{2n}{{C}_{n}} \right)$$$$$ D. \left( \dfrac{3n-5}{2} \right)\left( ^{2n}{{C}{n}} \right)$$$$$

Answer 1

We can write the given summation as $3\sum\limits_{n=0}^{n}{nC_{n}^{2}}-\sum\limits_{n=0}^{n}{C_{n}^{2}}$. We find the binomial expansion of $A={{\left( 1+x \right)}^{n}},B={{\left( 1+\dfrac{1}{x} \right)}^{n}}$ and then differentiate $A$to get$D$. We multiply respective sides of $A,B$ and equate the constant term to get$\sum\limits_{n=0}^{n}{C_{n}^{2}}$. We multiply respective sides of $A,B$ and equate coefficient of ${{x}^{-1}}$ to get$\sum\limits_{n=0}^{n}{nC_{n}^{2}}$.$$$$

Complete step-by-step solution
The given summation is
$\left( -1 \right)C_{0}^{2}+2C_{1}^{2}+5C_{2}^{2}+...+\left( 3n-1 \right)C_{n}^{2}$
We can write the above expression as a summation of the general term $\left( 3n-1 \right)C_{n}^{2}$ where $n=0,1,2,...n$. We have

& \left( -1 \right)C_{0}^{2}+2C_{1}^{2}+5C_{2}^{2}+...+\left( 3n-1 \right)C_{n}^{2}=\sum\limits_{n=0}^{n}{\left( 3n-1 \right)C_{n}^{2}} \\\ & \Rightarrow \left( -1 \right)C_{0}^{2}+2C_{1}^{2}+5C_{2}^{2}+...+\left( 3n-1 \right)C_{n}^{2}=3\sum\limits_{n=0}^{n}{nC_{n}^{2}}-\sum\limits_{n=0}^{n}{C_{n}^{2}}.....(1) \\\ \end{aligned}$$ We know that the binomial expansion of ${{\left( 1+x \right)}^{n}}$ is, $${{\left( 1+x \right)}^{n}}={{C}_{0}}+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}...+{{C}_{n}}{{x}^{n}}.....(2)$$ We know that the binomial of coefficient of ${{x}^{k}},k\in \left\\{ 0,1,2,...n \right\\}$ in above expression is${}^{n}{{C}_{k}}$. Let us substitute $\dfrac{1}{x}$ instead of $x$ to have, $${{\left( 1+\dfrac{1}{x} \right)}^{n}}={{C}_{0}}+{{C}_{1}}\dfrac{1}{x}+{{C}_{2}}\dfrac{1}{{{x}^{2}}}...+{{C}_{n}}\dfrac{1}{{{x}^{n}}}....(3)$$ We multiply equation (2) and (3) to have, $$\begin{aligned} & {{\left( 1+x \right)}^{n}}{{\left( 1+\dfrac{1}{x} \right)}^{n}}=C_{0}^{2}+{{C}_{0}}\left( x+\dfrac{1}{x} \right)+...+{{C}_{1}}{{C}_{0}}x+C_{1}^{2}... \\\ & \Rightarrow \dfrac{{{\left( 1+x \right)}^{2n}}}{{{x}^{n}}}=C_{0}^{2}+{{C}_{0}}\left( x+\dfrac{1}{x} \right)+...+{{C}_{1}}{{C}_{0}}x+C_{1}^{2}... \\\ \end{aligned}$$ We observe that the constant term in the in the right hand side of the above equation is$C_{0}^{2}+C_{1}^{2}+...+C_{n}^{2}$. The coefficient of ${{x}^{n}}$in the numerator ${{\left( 1+x \right)}^{2n}}$ of the fraction at the left hand side is $^{2n}{{C}_{n}}$. If we divide the term with coefficient of ${{x}^{n}}$ by ${{x}^{n}}$ as it exists in the denominator we will have the constant term at the left hand side which is $^{2n}{{C}_{n}}$. So we equate the coefficients of both side and have, $${}^{2n}{{C}_{n}}=C_{0}^{2}+C_{1}^{2}+...+C_{n}^{2}=\sum\limits_{n=0}^{n}{C_{n}^{2}}....(4)$$ Let us differentiate equation (2) with respect to $x$ and get $$n{{\left( 1+x \right)}^{n-1}}=0\cdot {{C}_{0}}+1\cdot {{C}_{1}}+2{{C}_{2}}x...+n{{C}_{n}}{{x}^{n-1}}.....(5)$$ We multiply respective sides of equation (3) and (5) to have, $$\begin{aligned} & n{{\left( 1+x \right)}^{n-1}}{{\left( 1+\dfrac{1}{x} \right)}^{n}}=\left( 0\cdot {{C}_{0}}+1\cdot {{C}_{1}}+2{{C}_{2}}x...+n{{C}_{n}}{{x}^{n-1}} \right)\left( {{C}_{0}}+{{C}_{1}}\dfrac{1}{x}+...+{{C}_{n}}\dfrac{1}{{{x}^{n}}} \right) \\\ & \Rightarrow \dfrac{n{{\left( 1+x \right)}^{2n-1}}}{{{x}^{n}}}=\left( 0\cdot {{C}_{0}}+1\cdot {{C}_{1}}+2{{C}_{2}}x...+n{{C}_{n}}{{x}^{n-1}} \right)\left( {{C}_{0}}+{{C}_{1}}\dfrac{1}{x}+...+{{C}_{n}}\dfrac{1}{{{x}^{n}}} \right) \\\ \end{aligned}$$ We observe that the coefficient of $\dfrac{1}{x}={{x}^{-1}}$ in the in the right hand side of the above equation is $0\cdot C_{0}^{2}+1\cdot C_{1}^{2}+2C_{2}^{2}...+nC_{n}^{2}$. The coefficient of ${{x}^{n-1}}$in the numerator ${{\left( 1+x \right)}^{2n-1}}$ of the fraction at the left hand side is $^{2n-1}{{C}_{n-1}}$. If we divide the term with coefficient of ${{x}^{n-1}}$ by ${{x}^{n}}$ as it exists in the denominator we will have the coefficient of ${{x}^{-1}}$ at the left hand side which is ${n^{2n - 1}}{C_{n - 1}}$. We equate the coefficients of both side and have, $$\begin{aligned} & n{}^{2n-1}{{C}_{n-1}}=0\cdot C_{0}^{2}+1\cdot C_{1}^{2}+2C_{2}^{2}...+nC_{n}^{2}=\sum\limits_{n=0}^{n}{nC_{n}^{2}} \\\ & \Rightarrow 3n{}^{2n-1}{{C}_{n-1}}=3\sum\limits_{n=0}^{n}{nC_{n}^{2}} \\\ & \Rightarrow 3\dfrac{n}{2}{}^{2n}{{C}_{n}}=3\sum\limits_{n=0}^{n}{nC_{n}^{2}}\left( \because {}^{2n}{{C}_{n}}=2{}^{2n-1}{{C}_{n-1}} \right)...(6) \\\ \end{aligned}$$ We put the values obtained the values from equation (4) and (6) in equation (1) to have, $$\begin{aligned} & \left( -1 \right)C_{0}^{2}+2C_{1}^{2}+5C_{2}^{2}+...+\left( 3n-1 \right)C_{n}^{2}=3\sum\limits_{n=0}^{n}{nC_{n}^{2}}-\sum\limits_{n=0}^{n}{C_{n}^{2}} \\\ & \Rightarrow \left( -1 \right)C_{0}^{2}+2C_{1}^{2}+5C_{2}^{2}+...+\left( 3n-1 \right)C_{n}^{2}=3\dfrac{n}{2}{}^{2n}{{C}_{n}}-{}^{2n}{{C}_{n}} \\\ & \Rightarrow \left( -1 \right)C_{0}^{2}+2C_{1}^{2}+5C_{2}^{2}+...+\left( 3n-1 \right)C_{n}^{2}=\left( \dfrac{3n-1}{2} \right){}^{2n}{{C}_{n}} \\\ \end{aligned}$$ So the correct option is B. **Note:** We also note that that coefficient of $k$ and $n-k$ are same in the binomial expansion of ${{\left( 1+x \right)}^{n}}$. We also note that the binomial expansion of ${{\left( 1+x \right)}^{n}}$ is specialized form of binomial expansion of ${{\left( x+y \right)}^{n}}$ which is given as$\left( ^{n}{{C}_{0}} \right){{x}^{n}}{{y}^{0}}+\left( ^{n}{{C}_{1}} \right){{x}^{n-1}}y+...+\left( ^{n}{{C}_{1}} \right){{x}^{0}}{{y}^{n}}$ where $n$ is always a non-negative integers and $x,y$ are real numbers.