AP 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
- A \(21\)
- B \(23\)
- C \(25\)
- D \(27\)
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…
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