7.Binomial Theorem
hard

$\left(2 x^2+\frac{1}{2 x}\right)^{11}$ ના વિસ્તરણમાં $x^{10}$ અને $x^7$ ના સહગુણકોનો નિરપેક્ષ તફાવત $........$ છે.

A

$12^3-12$

B

$11^3-11$

C

$10^3-10$

D

$13^3-13$

(JEE MAIN-2023)

Solution

$T _{ r +1}={ }^{11} C _{ r }\left(2 x ^2\right)^{11- r }\left(\frac{1}{2 x }\right)^{ T }$

$={ }^{11} C _{ r } 2^{11-2 r } x ^{22-3 r }$

$22-3 r =10 \quad \text { and } \quad 22-3 r =7$

$r =4 \quad \text { and } \quad r =5$

$\text { Coefficient of } x ^{10}={ }^{11} C _4 \cdot 2^3$

$\text { Coefficient of } x ^7={ }^{11} C _5 \cdot 2^1$

$\text { difference }={ }^{11} C _4 \cdot 2^3-{ }^{11} C _5 \cdot 2$

$=\frac{11 \times 10 \times 9 \times 8}{24} \times 8-\frac{11 \times 10 \times 9 \times 8 \times 7}{120} \times 2$

$=11 \times 10 \times 3 \times 8-11 \times 3 \times 4 \times 7$

$=11 \times 3 \times 4 \times(20-7)$

$=11 \times 12 \times 13$

$=12(12-1)(12+1)$

$=12\left(12^2-1\right)$

$=12^3-12$

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

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