The number of terms in the expansion of (y^{1 | vedclass

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

Question

(A) The general term of the expansion $(y^{1/5} + x^{1/10})^{55}$ is given by:$T_{r+1} = ^{55}C_{r} (y^{1/5})^{55-r} (x^{1/10})^{r}$$T_{r+1} = ^{55}C_

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) The general term of the expansion $(y^{1/5} + x^{1/10})^{55}$ is given by:$T_{r+1} = ^{55}C_{r} (y^{1/5})^{55-r} (x^{1/10})^{r}$$T_{r+1} = ^{55}C_{r} y^{\frac{55-r}{5}} x^{\frac{r}{10}}$For the powers of $x$ and $y$ to be free from radical signs,the exponents $\frac{55-r}{5}$ and $\frac{r}{10}$ must be integers.$\frac{55-r}{5} = 11 - \frac{r}{5}$ must be an integer,which implies $r$ must be a multiple of $5$.$\frac{r}{10}$ must be an integer,which implies $r$ must be a multiple of $10$.Combining these,$r$ must be a multiple of $10$ where $0 \le r \le 55$.Possible values for $r$ are $0, 10, 20, 30, 40, 50$.There are $6$ such values of $r$,so there are $6$ terms in the expansion that are free from radical signs.

The number of terms in the expansion of $(y^{1/5} + x^{1/10})^{55}$,in which powers of $x$ and $y$ are free from radical signs are

Similar Questions

Let the coefficients of powers of $x$ in the $2nd$, $3rd$, and $4th$ terms in the expansion of $(1+x)^{n}$, where $n$ is a positive integer, be in arithmetic progression. Then, the sum of the coefficients of odd powers of $x$ in the expansion is

The constant term in the expansion of $\left(\frac{x}{2}+\frac{1}{x}+\sqrt{2}\right)^5$ is $\frac{a \sqrt{2}}{2}$,then $a=$

The coefficient of $x^5$ in the expansion of $(x^2+2x+3)^5$ is

If the coefficients of $T_r, T_{r+1}, T_{r+2}$ terms of $(1+x)^{14}$ are in $A.P.$,then $r =$

If the coefficient of $a^{7} b^{8}$ in the expansion of $(a+2b+4ab)^{10}$ is $K \cdot 2^{16}$,then $K$ is equal to .... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module