The sum of all rational terms in the expansion of ( 2 5 1 + 5 3 1 ) 15 is equal to :

T r + 1 = 15 C r ( 5 3 1 ) r ( 2 5 1 ) 15 − r = 15 C r 5 3 r ⋅ 2 5 15 − r R = 3 λ , 15 μ ⇒ r = 0 , 15 2 rational terms ⇒ 15 C 0 2 3 + 15 C 15 ( 5 ) 5 = 8 + 3125 = 3133

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MathematicsBinomial TheoremIntegral and Fractional Part of a NumberMedium2 minPYQ_2024

MathematicsMediumsingle choice

The sum of all rational terms in the expansion of $(2^{\frac{1}{5}} + 5^{\frac{1}{3}})^{15}$ is equal to :

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Answer:

Solution:

$T_{r + 1} =^{15} C_{r} (5^{\frac{1}{3}})^{r} (\frac{1}{2 ^{5}})^{15 - r} =^{15} C_{r} 5^{\frac{r}{3}} \cdot 2^{\frac{15 - r}{5}} R = 3 λ, 15 μ \Rightarrow r = 0, 15$ 2 rational terms $\Rightarrow^{15} C_{0} 2^{3} +^{15} C_{15} (5)^{5} = 8 + 3125 = 3133$

Stream:JEESubject:MathematicsTopic:Binomial TheoremSubtopic:Integral and Fractional Part of a Number

⏱ 2mℹ️ Source: PYQ_2024

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