The numerically greatest term in the binomial | vedclass

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(B) Given expansion is $(2x - 3y)^5$ with $x = \frac{3}{2}$ and $y = \frac{2}{3}$.Let $T_{r+1}$ be the $(r+1)$-th term.$T_{r+1} = \binom{5}{r} (2x)^{5

Vedclass · Accepted Answer

$1080$

Vedclass · Answer

(B) Given expansion is $(2x - 3y)^5$ with $x = \frac{3}{2}$ and $y = \frac{2}{3}$.Let $T_{r+1}$ be the $(r+1)$-th term.$T_{r+1} = \binom{5}{r} (2x)^{5-r} (-3y)^r$.Substitute $x = \frac{3}{2}$ and $y = \frac{2}{3}$:$T_{r+1} = \binom{5}{r} (2 \cdot \frac{3}{2})^{5-r} (-3 \cdot \frac{2}{3})^r = \binom{5}{r} (3)^{5-r} (-2)^r$.We want the numerically greatest term,so we consider $|T_{r+1}| = \binom{5}{r} 3^{5-r} 2^r$.Calculate terms:$|T_1| = \binom{5}{0} 3^5 2^0 = 1 \cdot 243 \cdot 1 = 243$.$|T_2| = \binom{5}{1} 3^4 2^1 = 5 \cdot 81 \cdot 2 = 810$.$|T_3| = \binom{5}{2} 3^3 2^2 = 10 \cdot 27 \cdot 4 = 1080$.$|T_4| = \binom{5}{3} 3^2 2^3 = 10 \cdot 9 \cdot 8 = 720$.$|T_5| = \binom{5}{4} 3^1 2^4 = 5 \cdot 3 \cdot 16 = 240$.$|T_6| = \binom{5}{5} 3^0 2^5 = 1 \cdot 1 \cdot 32 = 32$.Comparing the values,the numerically greatest term is $1080$.

The numerically greatest term in the binomial expansion of $(2x - 3y)^5$ when $x = \frac{3}{2}$ and $y = \frac{2}{3}$ is

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