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prove statement by mathematical induction: 1+3+5+ ... + (2n-1)=n^2 I got up to Let S={n is N: 1+3+5+ ... + (2n-1) = n^2} Since 2(1) - 1 = 1^2 equation is true for n=1, thus 1 is S
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Find your basis... let n = 1 then we need to find P(k) next P(k+1)
I have to make S = N
1+3+5+ ... + (2k-1)=k^2 1+3+5+ ... + (2(k+1)-1)=(k+1)^2 1+3+5+ ... + (2k-1)=(k+1)(k+1) 1+3+5+ ... + (2k-1)=k^2+2k+1
I am getting there.. see that last line... to the right is your goal line.. your goal is to make the left look like k^2+2k+1
and I see it already if I do 1+3+5+...+(2k-1) + 2k+1 and subsitute that massive string with k^2
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Since 1+3+5+ ... + (2k-1)=k^2 k^2 +2k + 1 = k^2+2k+1
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