If x + y = 1 , then Σlimits_{r = 0}^n {{r^2}{,^n}{Cᵣ}{x^r}{y^{n

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7.Binomial Theorem

hard

If $x + y = 1$, then $\sum\limits_{r = 0}^n {{r^2}{\,^n}{C_r}{x^r}{y^{n - r}}} $ equals

$nxy$

$nx(x + yn)$

$nx(nx + y)$

None of these

Solution

$\sum\limits_{r = 0}^n {{r^2}{\,^n}{C_r}{x^r}{y^{n – r}}} $

$ = \sum\limits_{r = 0}^n {[r(r – 1) + r]{\,^n}} {C_r}{x^r}{y^{n – r}}$

$ = \sum\limits_{r = 0}^n {r(r – 1){\,^n}} {C_r}{x^r}{y^{n – r}} + \sum\limits_{r = 0}^n {{r^n}{C_r}{x^r}{y^{n – r}}} $

$ = \sum\limits_{r = 2}^{n – 2} {r(r – 1)\frac{n}{r}.\frac{{n – 1}}{{r – 1}}{\,^{n – 2}}{C_{r – 2}}{x^2}{x^{r – 2}}{y^{n – r}}} $ $ + \sum\limits_{r = 1}^{n – 1} {r\frac{n}{r}{\,^{n – 1}}{C_{r – 1}}x\,\,{x^{r – 1}}{y^{n – r}}} $

$ = n(n – 1){x^2}\sum\limits_{r = 2}^{n – 2} {{\,^{n – 2}}{C_{r – 2}}{x^{r – 2}}{y^{(n – 2) – (r – 2)}}} $ $+ nx\sum\limits_{r = 2}^{n – 1} {{\,^{n – 1}}{C_{r – 1}}{X^{r – 1}}{y^{(n – 1) – (r – 1)}}}$

$ = n(n – 1){x^2}{(x + y)^{n – 2}} + nx{(x + y)^{n – 1}}$

$ = n(n – 1){x^2} + nx,\,\,\,\,\,(\because x + y = 1)$

$ = nx(nx – x + 1) = nx(nx + y)\,,\,\,(\because x + y = 1)$

Standard 11

Mathematics

Let $(1+2 x)^{20}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$.Then $3 a_0+2 a_1+3 a_2+2 a_3+3 a_4+2 a_5+\ldots+2 a_{19}+3 a_{20}$ equals

normal

(KVPY-2009)

Let ${\left( {1 + x + {x^2}} \right)^{20}}\left( {2x + 1} \right) = {a_0} + {a_1}{x^1} + {a_2}{x^2} + … + {a_{41}}{x^{41}}$ , then $\frac{{{a_0}}}{1} + \frac{{{a_1}}}{2} + …. + \frac{{{a_{41}}}}{{42}}$ is equal to

normal

Let $[ x ]$ denote greatest integer less than or equal to $x .$ If for $n \in N ,\left(1-x+x^{3}\right)^{n}=\sum_{j=0}^{3 n} a_{j} x^{j}$, then $\sum_{j=0}^{\left[\frac{3 n}{2}\right]} a_{2 j}+4 \sum_{j=0}^{\left[\frac{3 n-1}{2}\right]} a_{2 j+1}$ is equal to

normal

(JEE MAIN-2021)

$2{C_0} + \frac{{{2^2}}}{2}{C_1} + \frac{{{2^3}}}{3}{C_2} + …. + \frac{{{2^{11}}}}{{11}}{C_{10}}$ = . . .

hard

If $\frac{{ }^{11} C_1}{2}+\frac{{ }^{11} C_2}{3}+\ldots . .+\frac{{ }^{11} C_9}{10}=\frac{n}{m}$ with $\operatorname{gcd}(n, m)=1$, then $n+m$ is equal to

hard

(JEE MAIN-2024)

If $x + y = 1$, then $\sum\limits_{r = 0}^n {{r^2}{\,^n}{C_r}{x^r}{y^{n - r}}} $ equals

Solution

Similar Questions

Let $(1+2 x)^{20}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$.Then $3 a_0+2 a_1+3 a_2+2 a_3+3 a_4+2 a_5+\ldots+2 a_{19}+3 a_{20}$ equals

Let ${\left( {1 + x + {x^2}} \right)^{20}}\left( {2x + 1} \right) = {a_0} + {a_1}{x^1} + {a_2}{x^2} + … + {a_{41}}{x^{41}}$ , then $\frac{{{a_0}}}{1} + \frac{{{a_1}}}{2} + …. + \frac{{{a_{41}}}}{{42}}$ is equal to

Let $[ x ]$ denote greatest integer less than or equal to $x .$ If for $n \in N ,\left(1-x+x^{3}\right)^{n}=\sum_{j=0}^{3 n} a_{j} x^{j}$, then $\sum_{j=0}^{\left[\frac{3 n}{2}\right]} a_{2 j}+4 \sum_{j=0}^{\left[\frac{3 n-1}{2}\right]} a_{2 j+1}$ is equal to

$2{C_0} + \frac{{{2^2}}}{2}{C_1} + \frac{{{2^3}}}{3}{C_2} + …. + \frac{{{2^{11}}}}{{11}}{C_{10}}$ = . . .

If $\frac{{ }^{11} C_1}{2}+\frac{{ }^{11} C_2}{3}+\ldots . .+\frac{{ }^{11} C_9}{10}=\frac{n}{m}$ with $\operatorname{gcd}(n, m)=1$, then $n+m$ is equal to

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