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\[\sum_{k=0}^{5}(3j+2)\] sum is equal to:______ Apparently the answer is 18j+12 How would I get that answer? Can anyone show me what to do, please? :O
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That's supposed to be "k=0" but the equal sign looks like a negative
Looks like a trick question. The summand (the thing that follows the giant sigma) does not involve any k. So, it is a constant as far as k is concerned...
So... if it is a constant, then... \[\huge \sum_{k=0}^n c=(n+1)c\]
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