practice problems pending from Q3

Q1. Use mathematical induction to derive the following formula for all n>=1:
1.(1!) + 2.(2!) + 3.(3!) .... n.(n!) = (n+1)! - 1
S1.
Pr. th.
(n+1)(n+1)! + (n+1)! - 1 = (n+2)! - 1
=> (n+1)!(n+2) - 1 = (n+2)! - 1
=> (n+2)! = (n+2)! H.P.

Q2.

+2(2!)+3(3!)+⋯+n(n!)=(n+1)


S2.

(a)

Assuming it holds for 'n', i.e.

1/1^2 ... + 1/n^2 <= 2 - 1/n


1/1^2 ... + 1/(n+1)^2 <= 2 - 1/(n+1)


If we can show that:

2 - 1/n + 1/(n+1)^2 <= 2 - 1/(n+1)

we are done.


If:

2 - 1/n + 1/(n+1)^2 <= 2 - 1/(n+1)

Then:

- 1/n + 1/(n+1)^2 <= - 1/(n+1)

then:

1/(n+1)^2 <= 1/n - 1/(n+1)

=>

1/(n+1)^2 <= 1/n(n+1)

which is true.

H.P.


(b)

Pr. th.

(n+1)/2^(n+1) + 2 - (n+2)/2^n = 2 - (n+3)/2^(n+1)

=> (n+1)/2^(n+1) - (n+2)/2^n = - (n+3)/2^(n+1)

=> (n+1)/2 - (n+2) = -(n+3)/2

=> n + 2 = (2n+4)/2

H.P. Q3.

S3.

Using the second principle of induction, assume that the given relation holds for n,n-1,n-2, then:


a_(n-1) = 5.2^(n-1) + 1

a_(n-2) = 5.2^(n-2) + 1

So

a_n = 3[5.2^(n-1) + 1] - 2[5.2^(n-2) + 1] = 3 - 2 + 15.2^(n-1) - 10.2^(n-2)

= 1 + 2^(n-1)[15 - 5] = 1 + 2^(n-1).10 = 1 + 5.2^n

H.P. Q4.

Use the Division Algorithm to establish the following:

(a) The square of any integer is either of the form 3k or 3k+1.

(b) The cube of any integer has one of the forms: 9k,9k+1, or 9k+8.

(c) The fourth power of any integer is either of the form 5k or 5k+1.

S4.

(a)

Any integer can be expresses as (3q + r)

(3q+r)^2 = 9q^2 + 6qr + r^2 = X

Case 1: r = 0 => X = 3k

Case 2: r = 1 => X = 3k + 1

Case 3: r = 4 => X = 3k + 4 = 1 + 3(k+1)

H.P.


(b)

(3q+r)^3 = 27q^3 + r^3 + 27q^2.r + 9q.r^2 = X

Case 1: r = 0 => X = 9k

r = 1 => X = 9k + 1

r = 2 => X = 9k + 8

H.P.


(c)

(5q + r)^2 = 25q^2 + r^2 + 10qr = X

r = 0 => X = 5k => X^2 = 5k

r = 1 => X = 5k + 1 => X^2 = 5k + 1

r = 2 => X = 5k + 4 => X^2 = 5k + 16 = 5k + 1

r = 3 => X = 5k + 9 => X^2 = 5k + 81 = 5k + 1

r = 4 => X = 5k + 16 = 5k + 1 => X^2 = 5k + 1

H.P.


Q5.


S5.

n^4+4n^2 + 11 = (n^2 + 2)^2 + 7 = X

n^2 = (2k+1)^2 = 4k^2 + 4k + 1

(n^2 + 2)^2 = [4k^2 + 4k + 3]^2

=>

X = 7 + 16k^4 + 16k^2 + 9 + 16k^3 + 24k + 24k^2 = 16 + 16k^4 + 16k^2 + 16k^3 + 24k + 24k^2

24k + 24k^2 = 24k(k+1)

k(k+1) is even => 24k(k+1) is a multiple of 16.

H.P. Q6.


S6.

u0 = 1/4

u1 = 3/8

u2 = 15/32

Assuming it converges, then:

L = 2L - 2L^2 => L = 2L^2 => L = 0,1/2


As it's increasing => L = 1/2

Let 1/2 - uk = vk(vk is a new sequence).

=> 1- 2u_k = 2v_k

=> 2u_k = 1 - 2v_k => uk = 1/2 - v_k


u_(k+1) = 1 - 2v_k - 2(1/2 - v_k)^2 = 1/2 - 2v_k^2 = 1/2 - v_(k+1) => v_(k+1) = 2v_k^2

v_0 = 1/2^2

v_1 = 1/2^3

v_2 = 1/2^5

v_3 = 1/2^9

...

pattern for 2's power in denominator: 2^0 + 1, 2^1 + 1, 2^2 + 1, 2^3 + 1

v_n = 1/2^[2^(2n)+1]

|u_k - L| = |L - u_k| = v_k = 1/2^[2^(2k)+1] <= 1/2^1000 => 2^(2k) + 1 >= 1000 => 2^2k >= 999

=> k >= 10 = answer

Q7.


S7.

1/a_n = 2/a_(n-1) - 1/a_(n-2)

=> 1/a_n = 1/a_(n-1) + 1/a_(n-1) - 1/a_(n-2)

=> 1/a_n is an A.P.

=>1/a_2019 = 1/a1 + (2019-1)(1/a2 - 1/a1) = 8075/3

=> p + q = 8078





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