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# Mathematical Induction?

Use mathematical induction to prove that:

(a) 2^2 + 4^2 +…+ (2n)^2 = [2n (n+1) (2n+1)]/3, for n ≥ 1, and

(b) 5n! ≥ 3^n, for n ≥ 2

Need full explanation of steps for revision.

### 2 Answers

- rajLv 71 decade agoFavorite Answer
LetP(n) denote the statement

1^2+2^2+3^2+......+n^2=n(n+1)(2n+1)/6

when n=1,P(1)=1(2)(3)/6=1 checks

let P(k) be true

i.e.1^2+2^2+3^2+.......+k^2=k(k+1)(2k+1)/6

to prove

P(k+1)=(k+1)(k+2)(2k+3)/6 is true

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

=>[1^2+2^2+3^2+......+k^2]+(k+1)^2

now 1^2+2^2+3^2+......+k^2=k(k+1)(2k+1)/6

substtituting

=k(k+1)(2k+1)/6 +(k+1)^2

=(k+1)[k(2k+1)+6(k+1)]/6

=(k+1)(2k^2+7k+6)/6

=(k+1)(k+2)(2k+3)/6

hence proved

2.try by your self

- 1 decade ago
Here is an outline of the steps.

(a) Basis: Show that it is true for n = 1. That's easy!

Induction: Assume true for n = k, and show true for n = k+1. Requires showing that

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

[ 2 k ( k + 1 ) ( 2k + 1 ) / 3 ] + [ 2 (k+1) ]²

That's also easy by multiplying by 3, dividing by k+1, etc.

(b) Basis: Show true for n = 2. Easy!

Induction: Assume 5k! ≥ 3^k, and show 5(k+1)! ≥ 3^(k+1).What is the connection between 5(k+1)! and 5k!? Multiplication by k+1, so

5k! ( k + 1 ) ≥ 3^k ( k + 1 ).

If you can now show that

3^k ( k + 1 ) ≥ 3^(k+1)

then the proof is finished.

Divide by 3^k, so must show

k + 1 ≥ 3

which is true because k ≥ 2