Let S = ∑ n = 1 ∞ ∑ r = 0 n r ( r n ) n 2 . Find S .

Since ∑ r = 0 n r ( r n ) = n 2 n − 1 , the summand is n / 2 n − 1 . Hence S = 1 + 1 + 4 3 + 2 1 + ⋯ = ∑ n = 1 ∞ n ( 1/2 ) n − 1 = ( 1 − 1/2 ) 2 1 = 4 .

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Mathematics - Binomial Theorem Question with Solution | TestHub

MathematicsBinomial TheoremProperties of Binomial coefficientsHard2 minQB

MathematicsHardnumerical

Let $S = \sum_{n = 1}^{\infty} \frac{n ^{2}}{\sum _{r = 0}^{n} r ( r n )}$ . Find $S$ .

Answer:

4.00

Solution:

Since $\sum_{r = 0}^{n} r (r n) = n 2^{n - 1}$ , the summand is $n / 2^{n - 1}$ . Hence $S = 1 + 1 + \frac{3}{4} + \frac{1}{2} + \dots = \sum_{n = 1}^{\infty} n (1/2)^{n - 1} = \frac{1}{( 1 - 1/2 ) ^{2}} = 4$ .

Stream:JEESubject:MathematicsTopic:Binomial TheoremSubtopic:Properties of Binomial coefficients

⏱ 2mℹ️ Source: QB

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