Question

If k‐th term of a sequence is ak = ( -1) ^k(2-k)k/2k-1

,what is its next term ak+1 when k is even?​

Answer 1

Given:

kth term of a sequence is

`a_k=((-1)^k(2-k)k)/(2k-1)`

To find:

The next term
`a_(k+1)` when k is even.

Solution:

We have,

`a_k=((-1)^k(2-k)k)/(2k-1)`

Put k=k+1, to get the next term.

`a_(k+1)=((-1)^(k+1)(2-(k+1))(k+1))/(2(k+1)-1)`

If k is even, then k+1 must be odd and odd power of -1 gives -1.

`a_(k+1)=((-1)(2-k-1)(k+1))/(2k+2-1)`

`a_(k+1)=((-1)(1-k)(1+k))/(2k+1)`

`a_(k+1)=-(1^2-k^2)/(2k+1)`
`[\because (a-b)(a+b)=a^2-b^2]`

`a_(k+1)=-(1-k^2)/(2k+1)`

Therefore, the next term is
`a_(k+1)=-(1-k^2)/(2k+1)`.