## Series Application / Induction Answers

1. The investment is worth \$11470 at the end of 5 years

2. There is \$24711 in the fund at the end of the 20 years.

3. a. The monthly repayments are \$1208.39

b. The total interest paid was \$140014

c. The outstanding amount was \$60305

4. a. S(n): 12 + 22 + ..     + n2 = 1/6.n(n + 1)(2n + 1)
S(1):   12 = 1/6 × 1 × 2 × 3      TRUE
∴ 1 S
Assume S(1) .. S(k) TRUE
∴ 12 + 22 + .. + k2 = 1/6.k(k + 1)(2k + 1)
Add (k + 1)2 to both sides
12 + 22 + .. + k2 + (k + 1)2 = 1/6.k(k + 1)(2k + 1) + (k + 1)2
Consider the right hand side of this expression
RHS = 1/6.(k + 1)[k(2k + 1) + 6(k + 1)]

=1/6(k + 1)[2k2 + 7k + 6]
=1/6(k + 1)(k + 2)(2k + 3)
= 1/6.n(n +1(2n + 1) with n = k + 1

∴ S(k + 1) also true
But S(1) true
∴ S(n) true for n positive integers

b. S(n): 13 + 23 + .. + (2n -1)3 = n2(2n2 − 1)
S(1):    13 = 12(2 × 12 − 1) TRUE
Assume S(1) to S(k) TRUE
∴ 13 + 23 + .. + (2k − 1)3 = k2(2k2 − 1)
Add (2k + 13) to both sides
Note: "next" to 2k − 1 is 2k + 1

∴ 13 + 23 + .. + (2k -1)3 + (2k + 1)3 = k2(2k2 − 1) + (2k + 1)3

Consider the right hand side:
RHS =2k4 − k2 + 8k3 + 12k2 + 6k + 1
Check (k + 1)2(2(k + 1)2 − 1)
= 2(k + 1)4 − (k + 1)2
= 2(k4 + 4k3 + 6k2 + 4k + 1) − (k2 + 2k + 1)
= 2k4 + 8k3 + 12k2 + 8k + 2 − k2 − 2k − 1
= 2k4 + 8k3 + 11k2 + 6k + 1
= RHS

∴ True for n = k + 1 if true for n = k.
But proved true for n = 1
∴ True for all positive integers n

c.

d. S(n):     32n − 1 = 4m       m is an integer
S(1):         32 − 1 = 8 = 4 × 2 ∴ True
Assume S(k) true i.e. 32k- 1 = 4m

Consider 32(k + 1) − 1 = 32.32k -1
=9(1 + 4m) − 1
=36m + 8
=4(9m + 2)
If m is an integer so is 9m + 2
∴ True for n = k + 1 if true for n = k
But true for n = 1 ∴ True for all n.

e. S(n): n3 + 6n2 + 2n = 3m, where m is an integer.
S(1):   9 = 3 × 3 TRUE
Assume k3 + 6k2+ 2k = 3m

Consider (k +1)3 + 6(k + 1)2 + 2(k + 1)
= k3 + 3k2 +3k + 1 + 6k2 + 12k + 6 + 2k + 2
= (k3 + 6k2 + 2k) + 3k2 + 3k + 1 + 12k + 6 + 2
= 3m + 3k2 + 15k + 9
=3(m + k2 + 5k + 3)

If m is an integer so is m + k2 + 5k + 3
∴ S(k + 1) true 1 + S(k) true
But S(1) true ∴ S(n) always true.

5. S(k) true. i.e. 1 + 3 + .. (2k − 1) = k2+ 4

Add 2k + 1 to both sides.
∴ 1 + 3 + ..+ (2k − 1) + (2k + 1) = k2 + 4 + 2k + 1
=(k + 1)2 + 4

∴ S(k + 1) true if S(k) true.
However S(1): 1 = 5 is FALSE

∴ We CANNOT deduce truth for all n.
In fact S(n) is NEVER true.

6. a. (1 + x)n(1 + x)n = ( nC0 + nC1 x + .. +nCnxn) . (nC0 + nC1x + .. + nCnxn)

Coefficient of xn is nC0 nCn + nC1nCn − 1 + nC2nCn − 2 + .. + nCn − 2nC2 + nCn − 1nC1 + nCnnC0

But nCk = nCn − k

∴ This can be simplified to nC02 + nC12 + .. nCn2
The coefficient of xn in (1 + x)2n is