math
posted by anonymous .
Evaluate the sum of the first 120 terms of the series whose nth term is equal to (4 n  14)

You can make use of:
Sum from n = 0 to N of n = 1/2 N (N+1) 
n.....1.....2.....3.....4.....5.....6
N....10...6....2.....2.....6.....10
The 120th term derives from L = a + (n  1)d where a = the first term, n = the number of terms and d = the common difference.
Then, the sum is S = n(a + L)/2.
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