The number of integral terms in the expansion | vedclass

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

Question

(B) The general term in the expansion of $(7^{1/2} + 11^{1/6})^{824}$ is given by $T_{r+1} = {}^{824}C_r (7)^{(824-r)/2} (11)^{r/6}$.For the term to b

Vedclass · Accepted Answer

$138$

Vedclass · Answer

(B) The general term in the expansion of $(7^{1/2} + 11^{1/6})^{824}$ is given by $T_{r+1} = {}^{824}C_r (7)^{(824-r)/2} (11)^{r/6}$.For the term to be an integer,the exponents of $7$ and $11$ must be integers.$1$. The exponent of $7$ is $(824-r)/2 = 412 - r/2$. For this to be an integer,$r$ must be an even number.$2$. The exponent of $11$ is $r/6$. For this to be an integer,$r$ must be a multiple of $6$.Combining these,$r$ must be a multiple of $	ext{lcm}(2, 6) = 6$.Thus,$r$ can take values $0, 6, 12, \dots, 822$.This is an arithmetic progression where $a = 0$,$d = 6$,and the last term $l = 822$.Using the formula $l = a + (n-1)d$,we get $822 = 0 + (n-1)6$.$n-1 = 822/6 = 137$.$n = 138$.Therefore,there are $138$ integral terms.

The number of integral terms in the expansion of $\{7^{(1/2)} + 11^{(1/6)}\}^{824}$ is equal to ...................

Similar Questions

If $\binom{p}{q} = {}^{p}C_{q}$ and $\sum_{i=0}^{m} \binom{10}{i} \binom{20}{m-i}$ is maximum,then $m=$

The $6^{th}$ term in the expansion of $\left( 2x^2 - \frac{1}{3x^2} \right)^{10}$ is

Find the coefficients of $x^r$ where $0 \le r \le (n - 1)$ in the expansion of $(x + 3)^{n - 1} + (x + 3)^{n - 2}(x + 2) + (x + 3)^{n - 3}(x + 2)^2 + ... + (x + 2)^{n - 1}$.

Find the $r^{\text{th}}$ term from the end in the expansion of $(x+a)^{n}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module