# series deffined by a(m+n+1)=-3/5a(n)+2 ,a1=2 find a formula for term of series and find limt of series

*print*Print*list*Cite

### 1 Answer

You need to consider the associated equation `x = -3/5x + 2` such that:

`x = -3/5x + 2 => x + 3/5x = 2 => 8x/5 = 2 => x = 10/8 => x = 5/4`

You need to form a new sequence such that:

`b_n = a_n - 5/4`

You need to evaluate the ratio `(b_(n+1))/(b_n)` such that:

`(b_(n+1))/(b_n) = (a_(n+1) - 5/4)/(a_n - 5/4)`

`(b_(n+1))/(b_n) = (-3/5*a_n + 2 - 5/4)/(a_n - 5/4)`

`(b_(n+1))/(b_n) = (-3/5*a_n + 3/4)/(a_n - 5/4) = -3/5`

You may evaluate the first term of the sequence `b_n` such that:

`b_1 = a_1 - 5/4`

Since the problem provides the value of `a_1` , hence, you need to substitute 2 for `a_1` such that:

`b_1 = 2 - 5/4 => b_1 = 3/4`

Hence, evaluating `b_n` yields:

`b_n = b_1*(-3/5)^(n-1) => b_n = (3/4)*(-3/5)^(n-1) `

`a_n = 5/4 + b_n => a_n = 5/4 + (3/4)*(-3/5)^(n-1) `

**Evaluating the limit of the sequence `a_n` yields:**

`lim_(n->oo) a_n = lim_(n->oo) (5/4 + (3/4)*(-3/5)^(n-1)) = 5/4`