Question

Determine the elements of the following sequence

1. A=n^2-3n+3 where: 4<n<8

2. B = n (-1) ^n where: 2<n<4

Answer 1

$A(5)=13, A(6)=21, A(7)=31. B(3)=-3.$ However I think the problem is not very correct. Actually may be it was meant, $A=\Sigma( n^2-3n+3).$ Hence $A= \Sigma n^2 -3\Sigma n+\Sigma3=\frac{n(n+1)(2n+1)}{6}-3\frac{n(n+1)}{2}+3n$ $=\frac{n(n^2-3n+5)}{3}.$ And B is given by $\Sigma (-1)^nn.$ For this let the sum be upto $2n$ terms. Then subtracting $\Sigma n$ we get the sum as $-2(1+3+\cdots +2n-1)=-2$$n^2$ . Hence required sum is $-2n^2 =B-\frac{2n(2n+1)}{2}\Rightarrow B=n$. If number of terms is $2n+1,$ then $B= n-(2n+1)=-n-1.$