{C_0}{C_r} + {C_1}{C_{r + 1}} + {C_2}{C_{r + | vedclass

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Question

(a) ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .... + {C_r}{x^r} + ....$ &hellip;..$(i)$

${\left( {1 + \frac{1}{x}} \right)^n} = {C_0} + {C_1}\fr

Vedclass · Accepted Answer

$\frac{{(2n)!}}{{(n - r)\,!\,(n + r)!}}$

Vedclass · Answer

(a) ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .... + {C_r}{x^r} + ....$ &hellip;..$(i)$

${\left( {1 + \frac{1}{x}} \right)^n} = {C_0} + {C_1}\frac{1}{x} + {C_2}\frac{1}{{{x^2}}} + ..... + {C_r}\frac{1}{{{x^r}}} + ....$ &hellip;..$(ii)$

Multiplying both sides and equating coefficient of ${x^r}$in $\frac{1}{{{x^n}}}{(1 + x)^{2n}}$or the coefficient of ${x^{n + r}}$in ${(1 + x)^{2n}}$ we get the value of required expression = $^{2n}{C_{n + r}} = \frac{{(2n)\,!}}{{(n - r)\,!\,(n + r)\,!}}$

Trick : Solving conversely.

Put $n = 1$ and $r = 0$ in first term , (given condition)

$(i)$ $^1{C_0}^1{C_0}{ + ^1}{C_1}^1{C_1} = 1 + 1 = 2$ ,    

$(r \le n)$ Put $n = 2,r = 1$, then

$(ii)$ $^2{C_0}^2{C_1}{ + ^2}{C_1}^2{C_2} = 2 + 2 = 4$

Now check the options

$(a)$ $(i)$ Put $n = 1,r = 0$, we get $\frac{{2!}}{{(1)!\,\,(1)!}} = 2$

$(ii)$ Put $n = 2,r = 1$, we get $\frac{{4!}}{{(1)\,!\,\,(3)\,!}} = 4$

${C_0}{C_r} + {C_1}{C_{r + 1}} + {C_2}{C_{r + 2}} + .... + {C_{n - r}}{C_n}$=

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