Question

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

Answer 1

Given:

kth term of a sequence is

$a_k=\dfrac{(-1)^k(2-k)k}{2k-1}$

To find:

The next term $a_{k+1}$ when k is even.

Solution:

We have,

$a_k=\dfrac{(-1)^k(2-k)k}{2k-1}$

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

$a_{k+1}=\dfrac{(-1)^{k+1}(2-(k+1))(k+1)}{2(k+1)-1}$

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

$a_{k+1}=\dfrac{(-1)(2-k-1)(k+1)}{2k+2-1}$

$a_{k+1}=\dfrac{(-1)(1-k)(1+k)}{2k+1}$

$a_{k+1}=-\dfrac{1^2-k^2}{2k+1}$    $[\because (a-b)(a+b)=a^2-b^2]$

$a_{k+1}=-\dfrac{1-k^2}{2k+1}$

Therefore, the next term is $a_{k+1}=-\dfrac{1-k^2}{2k+1}$.