If 5|b|

Question

(B) The given expression is $(2a + 5b)^{-4}$.We can rewrite this as $(2a)^{-4} \left(1 + \frac{5b}{2a}\right)^{-4}$,where $\left|\frac{5b}{2a}\right|

Vedclass · Accepted Answer

$-{ }^{6}C_{3} \frac{5^{3}}{2^{7}} \frac{b^{3}}{a^{7}}$

Vedclass · Answer

(B) The given expression is $(2a + 5b)^{-4}$.We can rewrite this as $(2a)^{-4} \left(1 + \frac{5b}{2a}ight)^{-4}$,where $\left|\frac{5b}{2a}ight| The general term for the expansion of $(1+x)^{-n}$ is $T_{r+1} = \frac{(-n)(-n-1)...(-n-r+1)}{r!} x^r$.For the $4^{th}$ term,$r = 3$ and $n = 4$.$T_{4} = (2a)^{-4} \left[ \frac{(-4)(-5)(-6)}{3!} \left(\frac{5b}{2a}ight)^3 ight]$.$T_{4} = \frac{1}{2^4 a^4} \left[ -\frac{4 	imes 5 	imes 6}{3 	imes 2 	imes 1} 	imes \frac{5^3 b^3}{2^3 a^3} ight]$.Using the combination formula ${ }^{n+r-1}C_{r} = \frac{(n+r-1)!}{r!(n-1)!}$,we note that $\frac{4 	imes 5 	imes 6}{3!} = { }^{6}C_{3}$.Thus,$T_{4} = -{ }^{6}C_{3} \frac{5^3 b^3}{2^{4+3} a^{4+3}} = -{ }^{6}C_{3} \frac{5^3 b^3}{2^7 a^7}$.

If $5|b| < 2|a|$,then the $4^{th}$ term in the expansion of $(2a + 5b)^{-4}$ is

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