Coefficient of x^{70} in x^2(1+x)^{98}+x^3(1+x)^{97}+ x^4(1+x)^{96}+ldots ldots . .+x^{54}(1+x)^{46}

English

Gujarati

Hindi

7.Binomial Theorem

hard

The coefficient of $x^{70}$ in $x^2(1+x)^{98}+x^3(1+x)^{97}+$ $x^4(1+x)^{96}+\ldots \ldots . .+x^{54}(1+x)^{46}$ is ${ }^{99} \mathrm{C}_p-{ }^{46} \mathrm{C}_{\mathrm{q}}$.

Then a possible value to $\mathrm{p}+\mathrm{q}$ is :

$55$

$61$

$68$

$83$

(JEE MAIN-2024)

Solution

$ x^2(1+x)^{98}+x^3\left(1+x^{97}\right)+x^4(1+x)^{96}+\ldots \ldots . $

$ x^{54}(1+x)^{46}$

Coeff. of $\mathrm{x}^{70}:{ }^{98} \mathrm{C}_{68}+{ }^{97} \mathrm{C}_{67}+{ }^{96} \mathrm{C}_{66}+\ldots \ldots \ldots$.

$ { }^{47} \mathrm{C}_{17}+{ }^{46} \mathrm{C}_{16} $

$ ={ }^{46} \mathrm{C}_{30}+{ }^{47} \mathrm{C}_{30}+\ldots \ldots \ldots . .{ }^{98} \mathrm{C}_{30} $

$ =\left({ }^{46} \mathrm{C}_{31}+{ }^{46} \mathrm{C}_{30}\right)+{ }^{47} \mathrm{C}_{30}+\ldots \ldots \ldots . .{ }^{98} \mathrm{C}_{30}-{ }^{46} \mathrm{C}_{31} $

$ ={ }^{47} \mathrm{C}_{31}+{ }^{47} \mathrm{C}_{30}+\ldots \ldots \ldots . .{ }^{98} \mathrm{C}_{30}-{ }^{46} \mathrm{C}_{31} $

$ \ldots \ldots $

$ ={ }^{99} \mathrm{C}_{31}-{ }^{46} \mathrm{C}_{31}={ }^{99} \mathrm{C}_{\mathrm{p}}-{ }^{46} \mathrm{C}_{\mathrm{q}}$

Possible values of $(p+q)$ are $62,83,99,46$

$\Rightarrow p+q=83$

Standard 11

Mathematics

Let $(1 + x)^m = C_0 + C_1x + C_2x^2 + C_3x^3 + . . . . . +C_mx^m$, where $C_r ={}^m{C_r}$ and $A = C_1C_3 + C_2C_4+ C_3C_5 + C_4C_6 + . . . . . .. + C_{m-2}C_m$, then which is false

normal

If the Coefficient of $x^{30}$ in the expansion of $\left(1+\frac{1}{x}\right)^6\left(1+x^2\right)^7\left(1-x^3\right)^8 ; x \neq 0$ is $\alpha$, then $|\alpha|$ equals

hard

(JEE MAIN-2024)

The coefficient of $x^{49}$ in the expansion of $(x – 1)$$\left( {x\, – \,\frac{1}{2}\,} \right)$$\left( {x\, – \,\frac{1}{{{2^2}}}\,} \right)$ …..$\left( {x\, – \,\frac{1}{{{2^{49}}}}\,} \right)$ is equal to

normal

$^n{C_0} – \frac{1}{2}{\,^n}{C_1} + \frac{1}{3}{\,^n}{C_2} – …… + {( – 1)^n}\frac{{^n{C_n}}}{{n + 1}} = $

normal

The coefficient of $x^r (0 \le r \le n – 1)$ in the expression :

$(x + 2)^{n-1} + (x + 2)^{n-2}. (x + 1) + (x + 2)^{n-3} . (x + 1)^2; + …… + (x + 1)^{n-1}$ is :

normal

The coefficient of $x^{70}$ in $x^2(1+x)^{98}+x^3(1+x)^{97}+$ $x^4(1+x)^{96}+\ldots \ldots . .+x^{54}(1+x)^{46}$ is ${ }^{99} \mathrm{C}_p-{ }^{46} \mathrm{C}_{\mathrm{q}}$. Then a possible value to $\mathrm{p}+\mathrm{q}$ is :

Solution

Similar Questions

Let $(1 + x)^m = C_0 + C_1x + C_2x^2 + C_3x^3 + . . . . . +C_mx^m$, where $C_r ={}^m{C_r}$ and $A = C_1C_3 + C_2C_4+ C_3C_5 + C_4C_6 + . . . . . .. + C_{m-2}C_m$, then which is false

If the Coefficient of $x^{30}$ in the expansion of $\left(1+\frac{1}{x}\right)^6\left(1+x^2\right)^7\left(1-x^3\right)^8 ; x \neq 0$ is $\alpha$, then $|\alpha|$ equals

The coefficient of $x^{49}$ in the expansion of $(x – 1)$$\left( {x\, – \,\frac{1}{2}\,} \right)$$\left( {x\, – \,\frac{1}{{{2^2}}}\,} \right)$ …..$\left( {x\, – \,\frac{1}{{{2^{49}}}}\,} \right)$ is equal to

$^n{C_0} – \frac{1}{2}{\,^n}{C_1} + \frac{1}{3}{\,^n}{C_2} – …… + {( – 1)^n}\frac{{^n{C_n}}}{{n + 1}} = $

The coefficient of $x^r (0 \le r \le n – 1)$ in the expression : $(x + 2)^{n-1} + (x + 2)^{n-2}. (x + 1) + (x + 2)^{n-3} . (x + 1)^2; + …… + (x + 1)^{n-1}$ is :

Start a Free Trial Now

The coefficient of $x^{70}$ in $x^2(1+x)^{98}+x^3(1+x)^{97}+$ $x^4(1+x)^{96}+\ldots \ldots . .+x^{54}(1+x)^{46}$ is ${ }^{99} \mathrm{C}_p-{ }^{46} \mathrm{C}_{\mathrm{q}}$.

Then a possible value to $\mathrm{p}+\mathrm{q}$ is :

The coefficient of $x^r (0 \le r \le n – 1)$ in the expression :

$(x + 2)^{n-1} + (x + 2)^{n-2}. (x + 1) + (x + 2)^{n-3} . (x + 1)^2; + …… + (x + 1)^{n-1}$ is :