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Using the principle of mathematical induction, prove each of the following for all n ϵ N:

1 + 2 + 3 + 4 + … + n = 1/2 n(n + 1)

2 + 4 + 6 + 8 + …. + 2n = n(n + 1)

2 + 6 + 18 + … + 2 × 3^{n–1} = (3^{n} –1)

1^{2} + 3^{2} + 5^{2} + 7^{2} + … + (2n – 1)^{2} =

1.2 + 2.2^{2} + 3.2^{3} + …. + n.2^{n} =(n – 1)2^{n+1} + 2.

3.2^{2} + 3^{2}.2^{3} + 3^{3}.2^{4} + …. + 3^{n}.2^{n+1} = (6^{n} – 1).

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= (n + 1)^{2}.

= (n + 1).

n × ( n + 1 ) × ( n + 2 ) is multiple of 6

(x^{2n} – y^{2n}) is divisible by (x + y).

(x^{2n} – 1) - 1 is divisible by (x – y), where x ≠ 1.

{(41)^{n} – (14)^{n}} is divisible by 27.

(4^{n} + 15n – 1) is divisible by 9.

(3^{2n+2} – 8n – 9) is divisible by 8.

(2^{3n} – 1) is a multiple of 7

3^{n}≥ 2^{n}.