The numerically greatest term in the expansio | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the expansion $(3x - 4y)^{23}$. Substituting $x = \frac{1}{6}$ and $y = \frac{1}{8}$,we get $(3(\frac{1}{6}) - 4(\frac{1}{8}))^{23} = (\frac

Vedclass · Accepted Answer

$^{23}C_{11} \cdot (\frac{1}{2})^{23}$

Vedclass · Answer

(A) Given the expansion $(3x - 4y)^{23}$. Substituting $x = \frac{1}{6}$ and $y = \frac{1}{8}$,we get $(3(\frac{1}{6}) - 4(\frac{1}{8}))^{23} = (\frac{1}{2} - \frac{1}{2})^{23} = 0^{23}$.However,for the numerically greatest term in $(a+b)^n$,we consider the absolute values $|T_{r+1}| = |^{n}C_r a^{n-r} b^r|$.Let $a = 3x = \frac{1}{2}$ and $b = -4y = -\frac{1}{2}$.$|T_{r+1}| = |^{23}C_r (\frac{1}{2})^{23-r} (-\frac{1}{2})^r| = ^{23}C_r (\frac{1}{2})^{23}$.To find the greatest term,we maximize $^{23}C_r$. For $n=23$,the maximum value of $^{23}C_r$ occurs at $r = \frac{n}{2} \pm 0.5$,i.e.,$r = 11$ and $r = 12$.Thus,the greatest terms are $T_{12}$ and $T_{13}$,both equal to $^{23}C_{11} (\frac{1}{2})^{23}$.

The numerically greatest term in the expansion of $(3x - 4y)^{23}$ when $x = \frac{1}{6}$ and $y = \frac{1}{8}$ is:

Similar Questions

Coefficient of $x$ in the expansion of ${\left( {{x^2} + \frac{a}{x}} \right)^5}$ is

If the expansion of $(y^2 + \frac{c}{y})^5$,the coefficient of $y$ is:

If the sum of the coefficients in the expansion of $(x - 2y + 3z)^n$ is $128$,then the greatest coefficient in the expansion of $(1 + x)^n$ is

The coefficient of $x^{70}$ in $x^2(1+x)^{98} + x^3(1+x)^{97} + x^4(1+x)^{96} + \ldots + x^{54}(1+x)^{46}$ is ${}^{99}C_p - {}^{46}C_q$. Then a possible value of $p+q$ is:

Let the coefficients of three consecutive terms in the binomial expansion of $(1+2x)^n$ be in the ratio $2:5:8$. Then the coefficient of the term,which is in the middle of these three terms,is $...........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module