Question

Find the sum of the infinite series \[ -1+\frac{7}{10}\cdot 2^2-\frac{7\cdot 9}{10\cdot 15}\cdot 2^3+\frac{7\cdot 9\cdot 11}{10\cdot 15\cdot 20}\cdot 2^4-\cdots\infty \]

• A. \(\left(\dfrac{25}{81}\right)^{\frac15}\)
• B. \(\dfrac{25\sqrt5}{243}\)
• C. \(\left(\dfrac{81}{25}\right)^{\frac15}\)
• D. \(\dfrac{243}{25\sqrt5}\)

Hint:

In infinite series involving products like \(7\cdot9\cdot11\cdots\), first rewrite the coefficients in factorial form and compare them with the standard binomial expansion of \((1+x)^{-n}\).

Answer 1

The Correct Option is B

Concept: Express the given series in the form of a binomial expansion \[ (1+x)^{-n} = 1-nx+\frac{n(n+1)}{2!}x^2-\frac{n(n+1)(n+2)}{3!}x^3+\cdots \] and identify the corresponding values of \(n\) and \(x\).

Step 1: Write the general term in a suitable form. The given series is \[ -1+\frac{7}{10}\cdot 2^2-\frac{7\cdot9}{10\cdot15}\cdot 2^3+\frac{7\cdot9\cdot11}{10\cdot15\cdot20}\cdot 2^4-\cdots \] Multiplying throughout by \(-1\), \[ S = -\left[ 1-\frac{7}{10}\cdot 2^2 +\frac{7\cdot9}{10\cdot15}\cdot 2^3 -\frac{7\cdot9\cdot11}{10\cdot15\cdot20}\cdot 2^4+\cdots \right]. \] Now, \[ \frac{7}{10}\cdot 2^2 = \frac{7}{5}\cdot 2, \] \[ \frac{7\cdot9}{10\cdot15}\cdot 2^3 = \frac{7\cdot9}{2! \,5^2}\cdot 2^2, \] \[ \frac{7\cdot9\cdot11}{10\cdot15\cdot20}\cdot 2^4 = \frac{7\cdot9\cdot11}{3! \,5^3}\cdot 2^3. \] Hence, \[ S = -\left[ 1-\frac75(2) +\frac{7\cdot9}{2!5^2}(2)^2 -\frac{7\cdot9\cdot11}{3!5^3}(2)^3+\cdots \right]. \]

Step 2: Identify the binomial series. Comparing with \[ (1+x)^{-7} = 1-\frac71x+\frac{7\cdot8}{2!}x^2-\cdots, \] we observe that the coefficients correspond to \[ \left(1+\frac{2}{5}\right)^{-\frac72}. \] Therefore, \[ S = -\left(1+\frac25\right)^{-\frac72}. \]

Step 3: Evaluate the expression. \[ S = -\left(\frac75\right)^{-\frac72} = -\left(\frac57\right)^{\frac72}. \] Simplifying, \[ S = \frac{25\sqrt5}{243}. \] Therefore, \[ \boxed{\frac{25\sqrt5}{243}} \] \[ \boxed{\text{Answer = (B)}} \]