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 :

A

$55$

B

$61$

C

$68$

D

$83$

(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

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.