Let a 1 , … , a n form an A.P. with partial sums s k and s 0 = 0 . Find ∑ k = 0 n ( k n ) s k .

With common difference d , s k = k a 1 + ( 2 k ) d . The binomial sums ∑ k ( k n ) = n 2 n − 1 and ∑ ( 2 k ) ( k n ) = ( 2 n ) 2 n − 2 give n a 1 2 n − 1 + d n ( n − 1 ) 2 n − 3 = 2 n − 2 ( n a 1 + s n ) .

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MathematicsSequence & SeriesA.P.Medium2 minQB

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Let $a_{1}, \dots, a_{n}$ form an A.P. with partial sums $s_{k}$ and $s_{0} = 0$ . Find $\sum_{k = 0}^{n} (k n) s_{k}$ .

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Solution:

With common difference $d, s_{k} = k a_{1} + (2 k) d$ . The binomial sums $\sum k (k n) = n 2^{n - 1}$ and $\sum (2 k) (k n) = (2 n) 2^{n - 2}$ give $n a_{1} 2^{n - 1} + d n (n - 1) 2^{n - 3} = 2^{n - 2} (n a_{1} + s_{n})$ .

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:A.P.

⏱ 2mℹ️ Source: QB

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