Find series: sum_(k=1)^n(3+4k)^2

reinzogoq

reinzogoq

Answered question

2022-09-03

Find series:
k = 1 n ( 3 + 4 k ) 2

Answer & Explanation

madirans2m

madirans2m

Beginner2022-09-04Added 17 answers

Solution:
k = 1 n ( 3 + 4 k ) 2 = k = 1 n ( 9 + 24 k + 16 k 2 ) = 9 k = 1 n 1 + 24 k = 1 n k + 16 k = 1 n k 2 = 9 × n + 24 × n ( n + 1 ) 2 + 16 n ( n + 1 ) ( 2 n + 1 6 = 9 n + 12 n ( n + 1 ) + 8 3 n ( n + 1 ) ( 2 n + 1 ) = n 3 [ 27 + 36 n + 36 + 8 ( 2 n 2 + 3 n + 1 ) ] = n 3 [ 71 + 60 n + 16 n 2 ] = n 3 ( 16 n 2 + 60 n + 71 )

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