TS EAMCET · Maths · Binomial Theorem
If sum of the coefficients of \(x^r(r=0,1,2, \ldots, 2 n)\) in the expansion of \(\left(1+3 x-2 x^2\right)^n\) is 128 , then \(\sum_{r=1}^{2 n} r \frac{(2 n)_{C_r}}{(2 n)_{C_{r-1}}}=\)
- A 120
- B 135
- C 90
- D 105
Answer & Solution
Correct Answer
(D) 105
Step-by-step Solution
Detailed explanation
Given sum of coefficient in expansion of \(\left(1+3 x-2 x^2\right)^n\) is 128 \(\therefore\) For sum of coefficient, putting \(n=1\) in the expression \(\left(1+3 x-2 x^2\right)^n\) \(\Rightarrow \quad(1+3 \cdot 1-2 \cdot 1)^n=128 \Rightarrow 2^n=2^7 \Rightarrow n=7\) Now…
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