The number of integral terms in the expansion | vedclass

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(B) The general term in the expansion of $(7^{1/3} + 11^{1/9})^{6561}$ is given by $T_{r+1} = {}^{6561}C_r (7^{1/3})^{6561-r} (11^{1/9})^r$.$T_{r+1} =

Vedclass · Accepted Answer

$730$

Vedclass · Answer

(B) The general term in the expansion of $(7^{1/3} + 11^{1/9})^{6561}$ is given by $T_{r+1} = {}^{6561}C_r (7^{1/3})^{6561-r} (11^{1/9})^r$.$T_{r+1} = {}^{6561}C_r \cdot 7^{(6561-r)/3} \cdot 11^{r/9}$.For the term to be an integer,the exponents of $7$ and $11$ must be integers.Thus,$(6561-r)/3$ must be an integer,which implies $r$ must be a multiple of $3$.Also,$r/9$ must be an integer,which implies $r$ must be a multiple of $9$.Since $9$ is a multiple of $3$,$r$ must be a multiple of $9$ such that $0 \le r \le 6561$.The possible values for $r$ are $0, 9, 18, \dots, 6561$.This is an arithmetic progression where $a = 0$,$d = 9$,and the last term $l = 6561$.Using the formula $l = a + (n-1)d$,we get $6561 = 0 + (n-1)9$.$6561/9 = n-1$ $\Rightarrow 729 = n-1$ $\Rightarrow n = 730$.Therefore,the number of integral terms is $730$.

The number of integral terms in the expansion of $(7^{1/3} + 11^{1/9})^{6561}$ is:

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