Question Number 188786 by Rupesh123 last updated on 07/Mar/23
Commented by Rupesh123 last updated on 07/Mar/23
Prove that:
Answered by mr W last updated on 07/Mar/23
![(1+x)^n (1+x)^n =(1+x)^(2n) [Σ_(k=0) ^n x^k ((n),(k) )][Σ_(r=0) ^n x^r ((n),(r) )]=(1+x)^(2n) [Σ_(k=0) ^n kx^(k−1) ((n),(k) )][Σ_(r=0) ^n x^r ((n),(r) )]+[Σ_(k=0) ^n x^k ((n),(k) )][Σ_(r=0) ^n rx^(r−1) ((n),(r) )]=2n(1+x)^(2n−1) [Σ_(k=0) ^n kx^(k−1) ((n),(k) )][Σ_(r=0) ^n x^r ((n),(r) )]=n(1+x)^(2n−1) [Σ_(k=0) ^n kx^(k−1) ((n),(k) )][Σ_(r=0) ^n x^r ((n),(r) )]=nΣ_(k=0) ^(2n−1) x^k (((2n−1)),(k) ) coef. of x^(n−1) : from RHS =n (((2n−1)),((n−1)) ) from LHS =Σ_(k=0) ^n k ((n),(k) ) ((n),((n−k)) )=Σ_(k=1) ^n k ((n),(k) )^2 ⇒Σ_(k=1) ^n k ((n),(k) )^2 =n (((2n−1)),((n−1)) )](https://www.tinkutara.com/question/Q188799.png)
$$\left(\mathrm{1}+{x}\right)^{{n}} \left(\mathrm{1}+{x}\right)^{{n}} =\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} \\ $$$$\left[\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{x}^{{k}} \begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\right]\left[\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{x}^{{r}} \begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\right]=\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} \\ $$$$\left[\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{kx}^{{k}−\mathrm{1}} \begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\right]\left[\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{x}^{{r}} \begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\right]+\left[\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{x}^{{k}} \begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\right]\left[\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{rx}^{{r}−\mathrm{1}} \begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\right]=\mathrm{2}{n}\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}−\mathrm{1}} \\ $$$$\left[\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{kx}^{{k}−\mathrm{1}} \begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\right]\left[\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{x}^{{r}} \begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\right]={n}\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}−\mathrm{1}} \\ $$$$\left[\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{kx}^{{k}−\mathrm{1}} \begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\right]\left[\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{x}^{{r}} \begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\right]={n}\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}{x}^{{k}} \begin{pmatrix}{\mathrm{2}{n}−\mathrm{1}}\\{{k}}\end{pmatrix} \\ $$$${coef}.\:{of}\:{x}^{{n}−\mathrm{1}} : \\ $$$${from}\:{RHS}\:={n}\begin{pmatrix}{\mathrm{2}{n}−\mathrm{1}}\\{{n}−\mathrm{1}}\end{pmatrix} \\ $$$${from}\:{LHS}\:=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{k}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\begin{pmatrix}{{n}}\\{{n}−{k}}\end{pmatrix}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}^{\mathrm{2}} \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}^{\mathrm{2}} ={n}\begin{pmatrix}{\mathrm{2}{n}−\mathrm{1}}\\{{n}−\mathrm{1}}\end{pmatrix} \\ $$
Commented by Rupesh123 last updated on 07/Mar/23
Nice solution, sir!