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evaluate sum?

25
E (4k-5)

k=1

2007-01-13 05:08:25 · 1 answers · asked by SICK MY DUCK! 1 in Science & Mathematics ➔ Mathematics

1 answers

25
E (4k-5)

k=1

>> Sum(4k - 5) from k=1 to k=25

= 4 [sum(k) from k=1 to k=25] - 5 [sum(1) from k=1 to k=25]
= 4 [25x26/2] - 5 [25] >>> 1 + 2 + .... + n = n(n+1)/2<<<
= 1300 - 125
= 1175

2007-01-13 05:30:44 · answer #1 · answered by Sheen 4 · 1⤊ 0⤋

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