TS EAMCET · Maths · Binomial Theorem
Let \(K\) be the number of rational terms in the expansion of \((\sqrt{2}+\sqrt[3]{3})^{6144}\). If the coefficient of \(x^{\mathrm{P}} \quad(\mathrm{P} \in \mathrm{N})\) in the expansion of \(\frac{1}{(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)}\) is \(\alpha_{\mathrm{P}}\), then \(\alpha_{\mathrm{K}}-\alpha_{\mathrm{K}+1}-\alpha_{\mathrm{K}-1}=\)
- A 1
- B 0
- C -2
- D 2
Answer & Solution
Correct Answer
(C) -2
Step-by-step Solution
Detailed explanation
\(T_{r+1} = \binom{6144}{r} (2)^{(6144-r)/2} (3)^{r/3}\). For rational terms, \(r\) must be a multiple of lcm(2, 3) = 6. \(0 \le r \le 6144 \implies r=6k\). \(0 \le 6k \le 6144 \implies 0 \le k \le 1024\). \(K = 1024 - 0 + 1 = 1025\).…
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