# Recurrence relations, compound interest, polynomials, number of combinations, & iteration

Please help with the following discrete math involving recurrence relations, compound interest, polynomials, number of combinations and iteration.

14. Individual membership fees at the evergreen tennis club were $50 in 1970 and have increased by $2 per year since then. Write a recurrence relation and initial conditions for the membership fee n years after 1970

2. Prove by mathematical induction that 4(2n) +3 is a solution to the recurrence relation sn = 2sn-1 -3 for n â‰¥ 1 with the initial condition s0 = 7

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Discrete Math: Recurrence relations, compound interest, polynomials, number of combinations, & iteration

14. Individual membership fees at the evergreen tennis club were $50 in 1970 and have increased by $2 per year since then. Write a recurrence relation and initial conditions for the membership fee n years after 1970

For one year, $50+$2

For two years $50+$4

For three years $50+$6

For four years $50+$8

Forwarding like this, we can see that fee for n years after 1970, is given by =$50+$2n

2. Prove by mathematical induction that 4(2n) +3 is a solution to the recurrence relation sn = 2sn-1 -3 for n â‰¥ 1 with the initial condition s0 = 7

We have sn =4(2n) +3, sn-1 =4(2n-1) +3

So, 2sn-1 -3=2[4(2n-1) +3]-3=4(2.2n-1) +3=4(2n) +3= sn.

So, sn=4(2n) +3, with s0=4+3=7 is the solution for the recurrence relation

Alternate method

Given that sn = 2sn-1 -3

Put fn = sn-1 -3. This means s0 = 7 implies f1 = s0 -3=4.

Hence the given relation can be written as sn-3 = 2sn-1 -6=2(sn-1 -3)

This implies fn =2fn-1

Clearly f1, f2, f3 ...form a G.P. with first term 4 and common ratio 2. So, fn =4.2n-1

Hence we have fn+1 = sn -3=4.2n.

This gives sn =4.2n+3

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