# mathematical induction sigma?

how do you solve using mathematical induction?

n

Σ i^5= n^2 (n+1)^2 (2n^2 + 2n - 1) divided by 12

i=1

### 1 Answer

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• 7 years ago
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Let's write this a little neater:

We have

1 + 32 + ... + n^5 = n²(n+1)²(2n² + 2n - 1)/12

Show that it's true for a base case: n = 1

LHS = 1^5 = 1

RHS = 1²(1+1)²(2*1² + 2*1 - 1)/12

= 1(2)²(3)/12 = 12/12 = 1.

Assume it to be true for n = k, i.e.

1 + 32 + ... + k^5 = k²(k+1)²(2k^2 + 2k - 1)/12

Prove it to be true for n = k + 1, in other words that

1 + 32 + ... + (k+1)^5 = (k+1)²([k+1] + 1)²(2[k+1]² + 2[k+1] - 1)/12

Before beginning, let's note the expansion for (k+1)^5 is

k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1

RHS = (k+1)²([k+1] + 1)²(2[k+1]² + 2[k+1] - 1)/12

= (k+1)²(k+2)²(2[k² + 2k + 1] + 2k + 2 - 1)/12

= (k² + 2k + 1)(k² + 4k + 4)(2k² + 4k + 2 + 2k + 1]/12

= (k² + 2k + 1)(k² + 4k + 4)(2k² + 6k + 3)/12

= (k^4 + 4k^3 + 4k² + 2k^3 + 8k² + 8k + k² + 4k + 4)(2k² + 6k + 3)/12

= (k^4 + 6k^3 + 13k² + 12k + 4)(2k² + 6k + 3)/12

= (2k^6 + 6k^5 + 3k^4 + 12k^5 + 36k^4 + 18k^3 + 26k^4 + 78k^3 + 39k² + 24k^3 + 72k² + 36k + 8k² + 24k + 12)/12

= (2k^6 + 18k^5 + 65k^4 + 120k^3 + 119k² + 60k + 12)/12

= (2k^6 + 6k^5 + 5k^4 - k^2)/12 + (12k^5 + 60k^4 + 120k^3 + 120k² + 60k

+ 12)/12

= k²(2k^4 + 6k^3 + 5k² - 1)/12 + (k^5 + 5k^4 + 10k^3 + 10k² + 5k + 1)

= k²(k² + 2k + 1)(2k² + 2k - 1)/12 + (k+1)^5

= k²(k+1)²(2k² + 2k - 1)/12 + (k+1)^5

= 1^5 + 2^5 + ... + k^5 + (k+1)^5 by assumption.

Proof complete.

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