Question #cb85a

2 Answers
Nov 15, 2017

#a_1=2, a_(n+1)=-5*a_n, n in NN.#

Explanation:

If we denote the terms of the sequence by #a_n, n in NN,# then,

#a_1=2.#

Note that, #a_2/a_1=a_3/a_2=a_4/a_3=...=-5.#

#:. a_2=-5*a_1,#

#a_3=-5*a_2,#

#a_4=-5*a_3,...#

Therefore, the recurrence formula for the sequence is given by,

#a_1=2, a_(n+1)=-5*a_n, n in NN.#

Nov 15, 2017

#a_n=2xx5^(n-1)xx(-1)^(n+1)#

Explanation:

Alternating positive negative is achieved by some #ul("variant")# of #(-1)^n# It is just a matter of making sure it is negative at the correct count by changing #n# in some way (if needed).

Let the #i^("th")# term be #a_i#

Note that #5^0=1#

#color(brown)("Dealing with just the numbers")#
#a_i=a_1=2 -> 2xx5^0#
#a_i=a_2=10->2xx5^1#
#a_i=a_3=50->2xx5xx5#
#a_i=a_4=250->2xx5xx5xx5#

So for any #i# the number is #a_i=2xx5^(i-1)#

THIS IS A GEOMETRIC SEQUENCE #->a_i=a_1xxr^(i-1)#

#color(brown)("Dealing with the sign")#

The first term is positive so #a_1=2xx5^0xx(-1)^1# is negative so no good.

To make the first term positive we can do this:

#a_1=2xx5^0xx(-1)^2color(white)("ddd")->color(white)("ddd")a_1=2xx5^0xx(+1) =2#

Thus the full expression for the #n^("th")# term is:

#a_n=2xx5^(n-1)xx(-1)^(n+1)#