Question

If $y = \frac{3}{4} + \frac{3 \cdot 5}{4 \cdot 8} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12} + ... \infty$, then

• A. $y^2 - 2y + 5 = 0$
• B. $y^2 + 2y - 7 = 0$
• C. $y^2 - 3y + 4 = 0$
• D. $y^2 + 4y - 6 = 0$

Hint:

For infinite series resembling binomial expansions, express the general term using factorials and test with $(1 + x)^n$. Solve resulting equations to find $n$ and $x$.

Answer 1

The Correct Option is B

The series is: \[ y = \sum_{n=1}^\infty \frac{3 \cdot 5 \cdot \ldots \cdot (2n+1)}{4 \cdot 8 \cdot \ldots \cdot (4n)} \] The general term is: \[ a_n = \frac{3 \cdot 5 \cdot \ldots \cdot (2n+1)}{4 \cdot 8 \cdot \ldots \cdot (4n)} = \frac{\prod_{k=1}^n (2k+1)}{\prod_{k=1}^n (4k)} \] Notice: \[ 2n+1 = 2(n+1) - 1, \quad \prod_{k=1}^n (2k+1) = \frac{(2n+2)!}{(2n+2) \cdot 2n \cdot \ldots \cdot 2} = \frac{(2n+2)!}{2^n (n+1)!} \] \[ 4n = 4 \cdot n, \quad \prod_{k=1}^n (4k) = 4^n \cdot n! \] Thus: \[ a_n = \frac{\frac{(2n+2)!}{2^n (n+1)!}}{4^n \cdot n!} = \frac{(2n+2)!}{2^n (n+1)! \cdot 4^n n!} = \frac{(2n+2)!}{2^n \cdot 4^n \cdot (n+1)! n!} = \frac{(2n+2)(2n+1)(2n)!}{2^{2n} \cdot 2^2 \cdot n! (n+1) n!} = \frac{(2n+1)!}{2^{2n+2} n! (n+1)!} \] Relate to binomial expansion $(1 + x)^n$. Compare with: \[ (1 + x)^n = \sum_{k=0}^\infty \binom{n}{k} x^k \] This series doesn’t directly match. Instead, assume $y$ corresponds to a binomial form. Test the series: \[ y = \sum_{n=1}^\infty \frac{\prod_{k=1}^n (2k+1)}{\prod_{k=1}^n (4k)} \] The original solution suggests $(1 + x)^n$. Try $n = -\frac{3}{2}$, $x = -\frac{1}{2}$: \[ (1 - \frac{1}{2})^{-\frac{3}{2}} = \left(\frac{1}{2}\right)^{-\frac{3}{2}} = 2^{\frac{3}{2}} = 2\sqrt{2} \] \[ y + 1 = 2\sqrt{2} \implies y = 2\sqrt{2} - 1 \] Check the quadratic: \[ y^2 + 2y - 7 = (2\sqrt{2} - 1)^2 + 2(2\sqrt{2} - 1) - 7 = 8 - 4\sqrt{2} + 1 + 4\sqrt{2} - 2 - 7 = 0 \] Option (2) is correct. Options (1), (3), and (4) do not satisfy $y = 2\sqrt{2} - 1$.