Induction (1 Viewer)

[conics2008]

Use mathematical induction to show that

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√1 + √2 + ・ ・ ・ + √n <=
4n+3/6 *√n


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[conics2008]

anyone..

just to avoid confusion its 4n+3 divide by 6 * (times) root of N

 

[Trebla]

Is it √1 + √2 + √3 + ........ + √n ≤ (4n + 3)√n / 6?
Because if the RHS is (4n + 3) / 6√n, then it would be false for n = 2 wouldn't it?

 

[Trebla]

I assume you mean:
√1 + √2 + √3 + ........ + √n ≤ (4n + 3)√n / 6

n = 1
LHS = 1
RHS = 7/6 > 1
LHS < RHS, thus true for n = 1
Assume statement is true for n = k
√1 + √2 + √3 + ........ + √k ≤ (4k + 3)√k / 6
Required to prove n = k + 1
√1 + √2 + √3 + ........ + √k + √(k + 1) ≤ (4k + 7)√[k+1] / 6
LHS = √1 + √2 + √3 + ........ + √k + √(k + 1)
≤ (4k + 3)√k / 6 + √(k + 1) by assumption
= {(4k + 3)√k + 6√(k + 1)} / 6
= √(k + 1) {(4k + 3)√k / √(k + 1) + 6} / 6
≤ (4k + 7)√[k+1] / 6 **
= RHS
**So we need to prove that (4k + 3)√k / √(k + 1) ≤ 4k + 1 to complete this step:
Working backwards and finding a starting point, I came up with this obviously fudged result lol: Someone help find a better one lol...
0 ≤ 1
Adding k(4k+3)² to both sides
k(4k+3)² ≤ k(4k+3)² + 1
Now k(4k+3)² + 1 = 16k³ + 24k² + 9k + 1
= (k + 1)(16k² + 8k + 1)
= (k + 1)(4k + 1)²
Thus: k(4k+3)² ≤ (k + 1)(4k + 1)²
square root both sides and divide √(k + 1) we get (4k + 3)√k / √(k + 1) ≤ 4k + 1 and that completes the step so insert conclusion...LOL

 

[3unitz]

√1 + √2 + √3 + ........ + √n ≤ (4n + 3)√n / 6

assume true n = k:
√1 + √2 + √3 + ........ + √k + √(k+1) ≤ [(4k + 3)√k / 6] + √(k+1)

n = k+1:
√1 + √2 + √3 + ........ + √k + √(k+1) ≤ [4(k+1) + 3]√(k+1) / 6

need to show:
[(4k + 3)√k / 6] + √(k+1) ≤ [4(k+1) + 3]√(k+1) / 6

(4k + 3)√k + 6√(k+1) ≤ 4k√(k+1) + 4√(k+1) + 3√(k+1)

(4k + 3)√k ≤ (4k + 1)√(k+1)

k(4k + 3)^2 ≤ (k+1)(4k + 1)^2

16k^3 + 24k^2 + 9k ≤ (k+1)(16k^2 + 8k + 1)

16k^3 + 24k^2 + 9k ≤ (16k^3 + 8k^2 + k) + (16k^2 + 8k + 1)

0 ≤ 1 true

therefore if true for n=k, true for n=k+1

 

[conics2008]

Thanks =]

Terbal I didn't get your workin out. I understood 3unitz.. I didn't know we can square, because that square root was pissin me off.

Thanks once again.