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Does anyone know how to write this in sigma notation? -2, + 1 + 4 + ... + 13 …
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please help!
anyone?
\[\sum_{n=0}^{5} (3n-2)\] is this you want?? check it please.
from the A.P. series, a=-2 and d=1-(-2)=3 and N=(l-a)/d=(13-(-2))/3=15/3=5 where a= first term, l=last term, N=total number of terms, d=common difference. so sum notation will be in general, \[\sum_{n=0}^{(l-a)/d}(a+nd)\]
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