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}

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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

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

normal

If the sum of the coefficients in the expansion of $(x+y)^{n}$ is $4096,$ then the greatest coefficient in the expansion is …. .

medium

(JEE MAIN-2021)

If $(1 + x) (1 + x + x^2) (1 + x + x^2 + x^3) …… (1 + x + x^2 + x^3 + …… + x^n)$

$\equiv a_0 + a_1x + a_2x^2 + a_3x^3 + …… + a_mx^m$ then $\sum\limits_{r\, = \,0}^m {\,\,{a_r}}$ has the value equal to

normal

Given $(1 – 2x + 5x^2 – 10x^3) (1 + x)^n = 1 + a_1x + a_2x^2 + ….$ and that $a_1^2\,= 2a_2$ then the value of $n$ is

normal

For natural numbers $m,n$ ,if ${\left( {1 – y} \right)^m}{\left( {1 + y} \right)^n} = 1 + {a_1}y + {a_2}{y^2} + \ldots \;$ and $a_1= a_2=10,$ then $(m,n)$ =______.

hard

(AIEEE-2006)

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

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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 :

If $(1 + x) (1 + x + x^2) (1 + x + x^2 + x^3) …… (1 + x + x^2 + x^3 + …… + x^n)$

$\equiv a_0 + a_1x + a_2x^2 + a_3x^3 + …… + a_mx^m$ then $\sum\limits_{r\, = \,0}^m {\,\,{a_r}}$ has the value equal to