If y = \; 3/4 + \; 3.5/4.8 + \; 3.5 · 7/4.8 · 12 + ·s ∞, then [AP EAPCET 24 May 2025 S1]

Correct Answer is : y2+2y−7=0\; y² + 2 y - 7 = 0y2+2y−7=0

Correct Answer is : y2+2y−7=0\; y² + 2 y - 7 = 0y2+2y−7=0

Test Index

Binomial Theorem

Section: Mathematics

Question : 2 of 33

Marks: +1, -0

y = \; \frac{3}{4} + \; \frac{3.5}{4.8} + \; \frac{3.5 \cdot 7}{4.8 \cdot 12} + \cdots \infty

Solution:

Given,

y = \; \frac{3}{4} + \; \frac{3 \cdot 5}{4 \cdot 8} + \; \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12} + \cdots + \infty

\; = \frac{3}{1!} \cdot \frac{1}{4} + \frac{3 \cdot 5}{2!} \cdot \frac{1}{4^2} + \frac{3 \cdot 5 \cdot 7}{3!} \cdot \frac{1}{4^3} + \cdots + \infty

Since, using binomial expansion

(1-x)^{-n} = 1 + n x + \frac{n(n+1)}{2!} x^2 + \frac{n(n+1)(n+2)}{3!} x^3 + \cdots

Here,

n x = \frac{3}{4} \;\;\; \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)

\; \frac{n(n+1)}{2} x^2 \; = \; \frac{3 \cdot 5}{2!} \cdot \frac{1}{4^2}

\Rightarrow \;\; n(n+1) x^2 = \frac{15}{16} \;\;\; \cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)

From Eq. (i) and (ii), we get

\; \frac{n(n+1) x^2}{n^2 x^2} \; = \; \frac{\frac{15}{16}}{\left( \frac{3}{4} \right)^2}

\; = \frac{\frac{15}{16}}{\frac{9}{16}} = \frac{5}{3}

\Rightarrow \; \frac{n+1}{n} = \frac{5}{3}

\Rightarrow 3 n + 3 = 5 n

\Rightarrow n = \frac{3}{2}

So,

x = \frac{3}{4} \cdot \frac{1}{n} = \frac{3}{4} \times \frac{2}{3} = \frac{1}{2}

\therefore \;\; y = (1-x)^{-n} - 1 = \left(1 - \frac{1}{2}\right)^{-\frac{3}{2}} - 1

\; = \left(\frac{1}{2}\right)^{-\frac{3}{2}} - 1 = 2^{\frac{3}{2}} - 1 = 2\sqrt{2} - 1

So,

y + 1 = 2\sqrt{2} - 1 + 1 = 2\sqrt{2}

(y+1)^2 = (2\sqrt{2})^2 = 8

\Rightarrow y^2 + 2 y + 1 - 8 = 0

\Rightarrow y^2 + 2 y - 7 = 0

Thus, the quadratic equation is

y^2 + 2 y - 7 = 0

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