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TS EAMCET · Maths · Binomial Theorem

If \(\quad \alpha=\frac{5}{2 ! 3}+\frac{5 \cdot 7}{3 ! 3^2}+\frac{5 \cdot 7 \cdot 9}{4 ! 3^3}+\ldots, \quad\) then \(\alpha^2+4 \alpha\) is equal to

  1. A \(21\)
  2. B \(23\)
  3. C \(25\)
  4. D \(27\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(23\)

Step-by-step Solution

Detailed explanation

Given that, \(\alpha=\frac{5}{2 ! 3}+\frac{5 \cdot 7}{3 ! 3^2}+\frac{5 \cdot 7 \cdot 9}{4 ! 3^3}+\ldots\) We know that, \[ \begin{aligned} (1+x)^n= & 1+\frac{n x}{1 !}+\frac{n(n-1)}{2 !} x^2 \\ & +\frac{n(n-1)(n-2)}{3 !} x^3+\ldots \end{aligned} \] On comparing Eqs. (i) and…