The number of terms which are free from radical signs in the expansion of ( y 1 5 + x 1 10 ) 55 is Options: A . &nbsp 5 B . &nbsp 6 C . &nbsp 7 D . &nbsp None of these

Answer: Option B : B In the expansion of ( y 1 5 + x 1 10 ) 55 , the general term is T r + 1 = 55 C r ( y 1 5 ) 55 − r ( x 1 10 ) r = 55 C r y 11 − r 5 x r 10 . This T r + 1 will be independent of radicals if the exponents r 5 and r 10 are integers, for 0 ≤ r ≤ 55 which is possible only when r = 0, 10, 20, 30, 40, 50. ∴ There are six terms viz. T 1 , T 11 , T 21 , T 31 , T 41 , T 51 which are independent of radicals.

The number of terms which are free from radical signs in the expansion of

Discussion Forum : Binomial Theorem

Question -

The number of terms which are free from radical signs in the expansion of $(y^{\frac{1}{5}} + x^{\frac{1}{10}})^{55}$ is

Options:

A . 5

B . 6

C . 7

D . None of these

Answer: Option B
:
B

In the expansion of $(y^{\frac{1}{5}} + x^{\frac{1}{10}})^{55}$ , the general

term is

$T_{r + 1}$ = $^{55} C_{r} (y^{\frac{1}{5}})^{55 - r} (x^{\frac{1}{10}})^{r}$ = $^{55} C_{r} y^{11 - \frac{r}{5}} x^{\frac{r}{10}}$ .

This $T_{r + 1}$ will be independent of radicals if the exponents $\frac{r}{5}$ and $\frac{r}{10}$ are integers, for

0 ≤ r ≤ 55 which is possible only when

r = 0, 10, 20, 30, 40, 50.

∴ There are six terms viz. $T_{1}, T_{11}, T_{21}, T_{31}, T_{41}, T_{51}$

which are independent of radicals.

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