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September 19th, 2016, 10:02 AM   #1
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induction

prove by induction for all positive integers n: 1+5+9+13+........+(4n-3)= n/2(4n-2)
i tried this by trying to prove n/2(4n-2)+ (4(k+1)-3) = k+1/2(4(k+1)-2) but it did not work out for me.
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September 19th, 2016, 10:16 AM   #2
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You mean the second step of the induction?

Assume $\displaystyle 1 + 5 + 9 + ... + (4k-3) = \frac{k(4k-2)}{2}$.

Then $\displaystyle 1 + 5 + 9 + ... + (4k-3) + [4(k+1)-3] = \frac{k(4k-2)}{2} + 4(k+1)-3 = \frac{k(4k-2) + 8(k+1) - 6}{2} = \frac{4k^2-2k + 8k+8 - 6}{2} = \frac{4k^2+6k+2}{2}=\frac{(k+1)(4k+2)}{2}=\frac{(k +1)(4(k+1)-2)}{2}$.
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September 19th, 2016, 10:51 AM   #3
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For the second step, you must say:

let k be an integer >=1 such that

1+5+...(4k-3)=.... (IH)

let's prove that

1+5+....(4k--3)+(4(k+1)-3)=......

then we use the induction hypothesis.
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