Question

s(n)=s(n-1)+2(n-1) with s(0)=3 , s(1)=1​

Answer 1

Answer࿐

Find two consecutive odd numbers such that two fifths of the smaller number exceeds two ninths of the larger by 4

Solution :

Let one odd number be ' 2n + 1 '

This is smallest odd number .

Other consecutive odd number be ' 2n + 3 '

This is largest odd number .

A/c , " Two fifths of the smaller number exceeds two ninths of the larger by 4 "

First consecutive smallest odd number :

= 2n + 1

= 2(12) + 1

= 24 + 1

= 25

Second consecutive largest odd number :

= 2n + 3

= 2(12) + 3

= 24 + 3

= 27

Alternative : You may solve this question by taking ' x ' as smallest consecutive odd number and ' x + 2 ' as biggest consecutive odd number .

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Answer 2

Answer:

solve

an=an−1+2an−2+2(1)

we can introduce bn−1=an. Then we get the linear recurrence

bn=bn−1+2bn−2(2)

Since

x2−x−2=(x−2)(x+1)(3)

we get the solution to the linear recurrence (2)

bn=c1(−1)n+c22n(4)

Therefore,

an=c1(−1)n+c22n−1(5)

Plugging the initial conditions, a0=0 and a1=2, into (5) we get

c1+c2−c1+2c2=1=3(6)

giving c1=−13,c2=43. Therefore,

an=−13(−1)n+432n−1(7)