1 - frac{3}{16} + frac{1 cdot 4}{1 cdot 2} le | vedclass

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Question

(B) The given series is $1 - \frac{3}{16} + \frac{1 \cdot 4}{2!} \left(\frac{3}{16}\right)^2 - \frac{1 \cdot 4 \cdot 7}{3!} \left(\frac{3}{16}\right)^

Vedclass · Accepted Answer

$\left(\frac{4}{5}\right)^{2/3}$

Vedclass · Answer

(B) The given series is $1 - \frac{3}{16} + \frac{1 \cdot 4}{2!} \left(\frac{3}{16}\right)^2 - \frac{1 \cdot 4 \cdot 7}{3!} \left(\frac{3}{16}\right)^3 + \ldots$ Comparing this with the binomial expansion $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots$ We have $nx = -\frac{3}{16}$ and $\frac{n(n-1)}{2!}x^2 = \frac{1 \cdot 4}{2!} \left(\frac{3}{16}\right)^2$. This implies $n(n-1)x^2 = 4 \left(\frac{3}{16}\right)^2$. Since $nx = -\frac{3}{16}$,we have $x = -\frac{3}{16n}$. Substituting $x$ in $n(n-1)x^2 = 4(nx)^2$,we get $n(n-1) \left(-\frac{3}{16n}\right)^2 = 4 \left(-\frac{3}{16}\right)^2$. $n(n-1) \frac{9}{256n^2} = 4 \cdot \frac{9}{256}$ $\Rightarrow \frac{n-1}{n} = 4$ $\Rightarrow n-1 = 4n$ $\Rightarrow 3n = -1$ $\Rightarrow n = -\frac{1}{3}$. Then $x = -\frac{3}{16(-1/3)} = \frac{9}{16}$. Thus,the sum is $(1+x)^n = (1 + \frac{9}{16})^{-1/3} = (\frac{25}{16})^{-1/3} = (\frac{16}{25})^{1/3} = ((\frac{4}{5})^2)^{1/3} = (\frac{4}{5})^{2/3}$.

$1 - \frac{3}{16} + \frac{1 \cdot 4}{1 \cdot 2} \left(\frac{3}{16}\right)^2 - \frac{1 \cdot 4 \cdot 7}{1 \cdot 2 \cdot 3} \left(\frac{3}{16}\right)^3 + \ldots$

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