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  • Number of terms free from radical sign in the expansion of (1 + 3 1/3 + 7 1/7 ) 10 is

Number of terms free from radical sign in the expansion of (1 + 3^1/3 + 7^1/7 )^10 is

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General term of (a_1+a_2+a_3.......a_r)^n=\frac{n!}{r_1!r_2!r_3!.......r_k!}a_1^{r_1}a_2^{r_2}a_3^{r_3}.......a_k^{r_k}

Now, General term of (1+3^{1/3}+7^{1/7})^{10}=\frac{10!}{a1b!c!}(1^{1})^{a}(3^{1/3})^{b}(7^{1/7})^{c}

Term free from radicle sign if

b is multiple of 3, c is multiple of 7 and we need a + b + c = 10

So there will be a total 5 combination possible, Hence, 5 terms which are free from radicle sign.

 

Posted by

himanshu.meshram

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Use Multinomial Theorem.

a+b+c=10 and for the terms to be rational (i.e.; free from radical sign) 6 combinations are possible.

So, the answer is 6.

Posted by

Atharva

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