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Write the sum using summation notation, assuming the suggested pattern continues. -10 - 2 + 6 + 14 + ... + 110
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First you can find the n value of the last term, 110, using the formula \[n ^{th}\ term= a + (n-1)d\] where the first term a = -10 and the common difference d = 8. By substituting in the formula we get \[110=-10+(n-1)8=-10+8n-8\ ............(1)\] When the terms in equation (1) are rearranged the value of n can be found from 8n = 128 ...............(2) Then the required summation can be found from: \[\sum_{110}^{-10}=\frac{n}{2}(-10+110)\]
\[\sum_{n=0}^{15}(-10+8n)\]
\[\sum_{n=0}^{\infty}(-10+8n)\]
is the asnwer the firstone
Yes, it is the first summation that you posted.
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