Write an expression to represent the sum of three times the square of a number and .
1 answer:
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Notice that
13 - 9 = 4
17 - 13 = 4
so it's likely that each pair of consecutive terms in the sum differ by 4. This means the last term, 149, is equal to 9 plus some multiple of 4 :
149 = 9 + 4k
140 = 4k
k = 140/4
k = 35
This tells you there are 35 + 1 = 36 terms in the sum (since the first term is 9 plus 0 times 4, and the last term is 9 plus 35 times 4). Among the given options, only the first choice contains the same amount of terms.
Put another way, we have
but if we make the sum start at k = 1, we need to replace every instance of k with k - 1, and accordingly adjust the upper limit in the sum.