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If $5|b| < 2|a|$. then the 4 th term in the expansion of $(2 a+5 b)^{-4}$ is
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Verified Answer
The correct answer is:
$-{ }^6 C_3 \frac{5^3}{2^7} \frac{b^3}{a^7}$
Given, binomial can be written as
$(2 a)^{-4}\left(1+\frac{5 b}{2 a}\right)^{-4}$, where $\left|\frac{5 b}{2 a}\right| < 1$
So, 4 th term will be $(2 a)^{-4}\left[\frac{(-4)(-5)(-6)}{3 !}\left(\frac{5 b}{2 a}\right)^3\right]$
$\left(\begin{array}{l}\because(1+x)^{-n} \\ =1+n x+\frac{n(n-1)}{2 !} x^2+ \\ \frac{n(n-1)(n-2)}{3 !} x^3+\ldots \infty \\ \text { Where, }|x| < 1\end{array}\right)$
$=\frac{1}{2^4 a^4}\left[(-1) \times \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 ! \times 3 \times 2 \times 1} \times \frac{5^3 b^3}{2^3 \cdot a^3}\right]$
$=-{ }^6 C_3 \cdot \frac{5^3 b^3}{2^7 \cdot a^7}$
$(2 a)^{-4}\left(1+\frac{5 b}{2 a}\right)^{-4}$, where $\left|\frac{5 b}{2 a}\right| < 1$
So, 4 th term will be $(2 a)^{-4}\left[\frac{(-4)(-5)(-6)}{3 !}\left(\frac{5 b}{2 a}\right)^3\right]$
$\left(\begin{array}{l}\because(1+x)^{-n} \\ =1+n x+\frac{n(n-1)}{2 !} x^2+ \\ \frac{n(n-1)(n-2)}{3 !} x^3+\ldots \infty \\ \text { Where, }|x| < 1\end{array}\right)$
$=\frac{1}{2^4 a^4}\left[(-1) \times \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 ! \times 3 \times 2 \times 1} \times \frac{5^3 b^3}{2^3 \cdot a^3}\right]$
$=-{ }^6 C_3 \cdot \frac{5^3 b^3}{2^7 \cdot a^7}$
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