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What is the sum of the series below?
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\[\sum_{i=1}^{6}(8i-2)\]
there are few enough terms here so you can just add
Use the formula: xn = a + d(n-1)
Where a is the first term and d is the common difference
n=number of terms
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replace \(i\) by 1, 2, 3, 4, 5, 6 and put plus signs between them \[8-2+8\times 2-2+8\times 3-2+8\times 4-2+8\times 5-2+8\times 6-2\]
or if you want to be fancy \[8(1+2+3+4+5+6)-2\times 6\]
or if you want to be real fancy, what @Luigi0210 said
Haha, what i did was not fancy at all.
thanks guys!
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yw
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