The sum of the rational terms in the expansion of -2-310 is

The correct option is C 41t_r+1=^10C_r.2^10 r/2.3^ r/5. This is rational if r/5and 5 r/2 are integers.Also, 0≤ r ≤ 10. Now, r must be a multiple f ...

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Standard XI

Mathematics

Sum of binomial coefficients with alternate signs

The sum of th...

Question

The sum of the rational terms in the expansion of ${(\sqrt{2} + \sqrt[5]{3})}^{10}$ is

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Solution

The correct option is C 41
$t_{r + 1} =^{1} 0 C_{r} {.2}^{\frac{10 - r}{2}} {.3}^{- \frac{r}{5}} . T h i s i s r a t i o n a l i f \frac{r}{5} a n d 5 - \frac{r}{2} a r e i n t e g e r s . A l s o, 0 \leq r \leq 10. N o w, r m u s t b e a m u l t i p l e f 10 (L C M o f 2 a n d 5) S o, r = 0, 10 ∴ T h e r a t i o n a l t e r m s a r e^{10} C_{0} {.2}^{5} {.3}^{0},^{10} C_{10} {.2}^{0} {.3}^{2} ∴ S u m o f r a t i o n a l t e r m s = 32 + 9 = 41$

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The sum of the rational terms in the expansion of (√2+5√3)10 is

The correct option is C 41tr+1=10Cr.210−r2.3−r5. This is rational if r5and 5−r2 are integers.Also, 0≤r≤10.Now, r must be a multiple f 10 (LCM of 2 and 5)So, r=0, 10∴The rational terms are 10C0.25.30,10C10.20.32∴ Sum of rational terms=32+9=41

The sum of the rational terms in the expansion of ${(\sqrt{2} + \sqrt[5]{3})}^{10}$ is