If $\left(x-2+\frac{1}{x}\right)^{30}=n_0 x^{30}+n_1 x^{29}+\ldots \ldots+n_{29} x+n_{30}+n_{31} x^{-1}+\ldots \ldots+n_{60} x^{-30}$ and $C=n_0+n_1+n_2+\ldots \ldots+n_{60}$. Find the value of $(a+b)$ if $C-n_{30}=-\left(\begin{array}{l}a \\ b\end{array}\right)$. $\left[{\left(\begin{array}{l}n \\ r\end{array}\right) \text{ denotes }{ }^n C_r}\right]$

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Medium

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Application +4 | -1

\left(x-2+\frac{1}{x}\right)^{30}=n_0 x^{30}+n_1 x^{29}+\ldots \ldots+n_{29} x+n_{30}+n_{31} x^{-1}+\ldots \ldots+n_{60} x^{-30}

and

C=n_0+n_1+n_2+\ldots \ldots+n_{60}

. Find the value of

(a+b)

C-n_{30}=-\left(\begin{array}{l}a \\ b\end{array}\right)

\left[{\left(\begin{array}{l}n \\ r\end{array}\right) \text{ denotes }{ }^n C_r}\right]

View answer and explanation

Correct option: A

Option A (90) is correct because the sum of coefficients C is 0, and C - n_30 = - (a choose b), leading to a = 60, b = 30, and a+b = 90.