M2 Maths Questions

1 (a) Assume that when n=k P(k) is true

1+2+2^2+...+2^(k-1)=2^k-1

when n=k+1
L.H.S.
=1+2+2^2+...+2^(k-1)+2^k
=2^k-1+2^k
=2^(k+1)-1
=R.H.S.

So, P(k+1) is true.

(b)when n=1 L.H.S.=1 R.H.S.=2^1-1=1

By M.I, for all positive values of n P(n) is true.

2.(a) P(n) be ‘1^2 + 3^2 + 5^2 +… (2n-1)^2=(1/3)n{(4n^2)-1}

So, for n=k+1, we need to add [2(k+1)-1]^2=(2k+1)^2 on both sides.

(b) P(n) be ‘(1+n)+(2+n)+(3+n)+…+(n+n)=(1/2)n(3n+1) ’

P(k)=(1+k)+(2+k)+(3+k)+…+(k+k)
P(k+1)=(1+k+1)+(2+k+1)+(3+k+1)+...+(k+1+k+1)

So, for n=k+1, we need to add 1+1+...+1+(k+1+k+1)=2k+k+2=3k+2 on both sides

3 when n=1, (2n+1)(2n+3)(2n+5)=3*5*7 which is divisible by 3
Assume that 3 | (2k+1)(2k+3)(2k+5) or (2k+1)(2k+3)(2k+5)=3M
when n=k+1, (2k+3)(2k+5)(2k+7) =2k(2k+3)(2k+5)+7(2k+3)(2k+5)
=2k(2k+3)(2k+5)+(2k+3)(2k+5)+6(2k+3)(2k+5)
=3M+6(2k+3)(2k+5)
which is divisible by 3

by mathematical induction, that for all positive integers n, (2n+1)(2n+3)(2n+5) is divisible by 3

1.(a) Let P(n) be the statement: 1+2+2^2+...+2^(n -1)=2^n-1

For n = 1, LHS = 1
RHS = 2^1 - 1 = 1
Therefore, P(1) is true

Assume P(k) is true
i.e. 1+2+2^2+...+2^(k -1)=2^k-1

For n = k+1
1+2+2^2+...+2^(k -1) + 2^k = 2^k-1 + 2^k
= 2^(k+1) +1
Therefore P(k+1) is also true

(b) From (a), by the principle of M.I., P(n) is true for all positive intergers n.


2. (a) Assume P(k) is true
i.e. 1^2 + 3^2 + 5^2 +… (2k-1)^2 = (1/3)k{(4k^2)-1}

For n = k +1,
1^2+3^2+5^2+(2k-1)^2+(2k+1)^2=(1/3)k{(4k^2)-1} + (2k+1)^2

Therefore, The expression is [2(k+1) - 1]^2 = (2k + 1)^2

(b)
‘(1+n)+(2+n)+(3+n)+…+(n+n)=(1/2)n(3n+1)
Final expression:
Put n = k+1

(2 + k) + (3+k) + (4+k) + ... + (2k+2) = (1/2)(k+1)[3(k+1) +1]

So the RHS is (1/2)(k+1)[3(k+1) +1] = (k+1)(3k+4) / 2


3. Let P(n) be the statement: (2n+1)(2n+3)(2n+5) is divisible by 3

For n = 1, (2x1 + 1)(2x1+3)(2x1+5) = 105 which is divisible by 3

Assume P(k) is true
i.e. (2k+1)(2k+3)(2k+5) = 3N where N is an interger

For n = k+1,
(2k+3) (2k+5) (2k +7) = 2k (2k+3)(2k+5) + 6(2k+3)(2k+5) + (2k+3)(2k+5)
= (2k+1)(2k+3)(2k+5) + 6(2k+3)(2k+5)
= 3N + 6(2k+3)(2k+5)
3[N + (2k+3)(2k+5)] which is divisible by 3
Therefore, P(k+1) is also true
By the principle of M.I., P(n) is true for all positive integers n.