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Binomial Theorem question

2023 · 1 Feb · Shift 2 · Q45

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Mathematics
Binomial Theorem
2023 · 1 Feb · Shift 2 · Q45

Binomial Theorem question

2023 · 1 Feb · Shift 2 · Q45

JEE MainMathematicsBinomial TheoremNumerical+4 / −1

If the term without

x

in the expansion of

\left(x^{\frac{2}{3}}+\frac{\alpha}{x^{3}}\right)^{22}

is 7315 , then

|\alpha|

is equal to

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 1

Consider the general term in the expansion of $\left(x^{\frac{2}{3}}+\frac{\alpha}{x^3}\right)^{22}.$

Using the binomial theorem, the general term is $T_{r+1}=\binom{22}{r}\left(x^{\frac23}\right)^{22-r}\left(\alpha x^{-3}\right)^r.$

Simplify the powers: $T_{r+1}=\binom{22}{r}\alpha^r x^{\frac{2}{3}(22-r)-3r}.$

So the exponent of $x$ is $\frac{44}{3}-\frac{2r}{3}-3r=\frac{44}{3}-\frac{11r}{3}=\frac{44-11r}{3}.$

For the term independent of $x$ , this exponent must be $0$ : $\frac{44-11r}{3}=0$ $44-11r=0$ $r=4.$
Hence the constant term is the $5$ th term: $T_5=\binom{22}{4}\alpha^4.$

Given that this term equals $7315$ , $\binom{22}{4}\alpha^4=7315.$

Now, $\binom{22}{4}=\frac{22\cdot21\cdot20\cdot19}{4\cdot3\cdot2\cdot1}=7315.$

Therefore, $7315\alpha^4=7315$ $\alpha^4=1.$

Thus $|\alpha|=1.$

So the required integer is $\boxed{1}.$

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