For n ≥ 0 , define a n = ( ln 10 ) n ∑ k = 0 n [ 1/ ( k ! ( n − k )!)] . Find ∑ n = 0 ∞ a n .

The inner sum is 2 n / n ! . Therefore ∑ a n = exp ( 2 ln 10 ) = 100 .

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For $n \geq 0$ , define $a_{n} = (ln 10)^{n} \sum_{k = 0}^{n} [1/ (k! (n - k)!)]$ . Find $\sum_{n = 0}^{\infty} a_{n}$ .

Answer:

Solution:

The inner sum is $2^{n} / n!$ . Therefore $\sum a_{n} = exp (2 ln 10) = 100$ .

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Special sequences/series

⏱ 2mℹ️ Source: QB

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