The sigma for the given series is given as
[tex]\sum ^{\infty}_{n\mathop=1}=(n+1)^{2\text{ }}(-1)^{n+1}[/tex]Above is the required sigma notation for the question.
Now, let's use it to find the first term 4, ( where n=1 )
[tex]\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}[/tex]You can also get -9 ( the second term 0 by substituting in n=2. etc.
