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Binomial Theorem question

2023 · 8 Apr · Shift 2 · Q28

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Mathematics
Binomial Theorem
2023 · 8 Apr · Shift 2 · Q28

Binomial Theorem question

2023 · 8 Apr · Shift 2 · Q28

JEE MainMathematicsBinomial TheoremMCQ+4 / −1

The absolute difference of the coefficients of

x^{10}

and

x^{7}

in the expansion of

\left(2 x^{2}+\frac{1}{2 x}\right)^{11}

is equal to :

A
$11^{3}-11$
B
$13^{3}-13$
C
$12^{3}-12$
D
$10^{3}-10$

View written solutionFree

Correct answer: C

General term of the binomial expansion

For $\left(2x^2+\frac{1}{2x}\right)^{11},$ the general term is $T_{r+1}=\binom{11}{r}(2x^2)^{11-r}\left(\frac{1}{2x}\right)^r.$

Now simplify: $T_{r+1}=\binom{11}{r}2^{11-r}x^{2(11-r)}\cdot 2^{-r}x^{-r}$ $=\binom{11}{r}2^{11-2r}x^{22-3r}.$

So, the coefficient of the term containing $x^{22-3r}$ is $\binom{11}{r}2^{11-2r}.$

Coefficient of $x^{10}$

We need: $22-3r=10$ $3r=12 \Rightarrow r=4.$

Thus the coefficient of $x^{10}$ is $\binom{11}{4}2^{11-8}=\binom{11}{4}2^3.$

Now, $\binom{11}{4}=330,$ so the coefficient is $330\times 8=2640.$

Coefficient of $x^7$

We need: $22-3r=7$ $3r=15 \Rightarrow r=5.$

Thus the coefficient of $x^7$ is $\binom{11}{5}2^{11-10}=\binom{11}{5}2.$

Now, $\binom{11}{5}=462,$ so the coefficient is $462\times 2=924.$

Absolute difference

$|2640-924|=1716.$

Match with options

Check each:

A: $11^3-11=1331-11=1320$
B: $13^3-13=2197-13=2184$
C: $12^3-12=1728-12=1716$
D: $10^3-10=1000-10=990$

Thus, $1716=12^3-12.$

So the correct option is C.

Comparison with stored answer

Stored correct answer: C

This matches our derived answer.

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