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Mathematics Question on Binomial theorem

The numerically greatest term in the expansion of (3−5x)11(3-5x)^{11} when x=15,x = \frac{1}{5}, is

A

55×3955 \times 3^9

B

55×3655 \times 3^6

C

45×3945 \times 3^9

D

45×3645 \times 3^6

Answer

55×3955 \times 3^9

Explanation

Solution

we have
(3−5)11=311(1−5x3)11\left(3-5\right)^{11} = 3^{11} \left(1 - \frac{5x}{3}\right)^{11}
= 3^{11} \left(1- \frac{5}{3}. \frac{1}{5}\right)^{11} \left\\{\because x = \frac{1}{5}\right\\}
=311(1−13)11= 3^{11} \left(1- \frac{1}{3}\right)^{11}
Now, r=∣x∣(n+1)∣x∣+1 r= \frac{\left|x\right|\left(n+1\right)}{\left|x\right|+1}
=∣−13∣(11+1)∣−13∣+1= \frac{\left|- \frac{1}{3}\right|\left(11+1\right)}{\left|- \frac{1}{3}\right| + 1}
=443= \frac{4}{\frac{4}{3}}
⇒r=3\Rightarrow r = 3
Therefore, 3rd(T3)3rd (T_3) and (3+1)=4th(T4)(3 + 1) = 4th (T_4) terms are numerically greatest in the expansion of (3−5x)11(3 - 5x)^{11}.
Hence, greatest term = T3T_3
=311∣11C2(1)9(−13)2∣=3^{11} \left|^{11}C_{2} \left(1\right)^{9} \left(- \frac{1}{3}\right)^{2} \right|
=311∣11×101.2.9∣=3^{11} \left|\frac{11\times10}{1.2.9}\right|
=55×39= 55 \times3^{9}
and T4=311∣11C3(1)8(−13)3∣T_{4} =3^{11}\left|^{11}C_{3} \left(1\right)^{8} \left(- \frac{1}{3}\right)^{3}\right|
=311∣11×10×91.2.3.(−127)∣=55×39= 3^{11} \left|\frac{11 \times10\times9}{1.2.3} .\left( - \frac{1}{27}\right)\right| = 55\times3^{9}
Hence Greatest term (numerically) =T3=T4 = T_{3} =T_{4}
=55×39= 55 \times3^{9}