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4-9+16 - 25+... written as sigma notation; n=1

User Brtgmaden
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1 Answer

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The sigma for the given series is given as


\sum ^(\infty)_(n\mathop=1)=(n+1)^{2\text{ }}(-1)^(n+1)

Above is the required sigma notation for the question.

Now, let's use it to find the first term 4, ( where n=1 )


\begin{gathered} \sum ^(\infty)_(n\mathop=1)=(1+1)^2(-1)^(1+1) \\ =2^2(-1)^2 \\ =2^{2\text{ }}(1)=2^2\text{ = 4} \end{gathered}

You can also get -9 ( the second term 0 by substituting in n=2. etc.

User Ryan Chouinard
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