So we assume
1+5+9+....(4k-3) = k(2k-1).
What we want to show is that if we substitute k+1 for n in
1+5+9+...(4n-3) = n(2n-1)
It will also work.
That is we want to show that based on the assumption that it
works for k, that we'll end up with this without the question
mark over the equal sign:
1+5+9+...+[4(k+1)-3] ≟ (k+1)[2(k+1)-1]
The question mark above the equal sign is to show that we
have not shown that yet. The above is what we must show.
We will simplify what we are to show:
1+5+9+...+[4k+4-3] ≟ (k+1)[2k+2-1]
1+5+9+...+(4k+1) ≟ (k+1)(2k+1)
1+5+9+...+(4k+1) ≟ 2k²+3k+1
Now we will show it. We start with:
1+5+9+...+(4k-3) = k(2k-1)
We add the k+1st term, which is (4k+1), to both sides,
and hope that the right side comes out to to the
expression above, which is 2k²+3k+1:
1+5+9+...+(4k-3)+(4k+1) = k(2k-1)+(4k+1)
1+5+9+...+(4k-3)+(4k+1) = 2k²-k+4k+1
1+5+9+...+(4k-3)+(4k+1) = 2k²+3k+1
So we have shown that if the formula works for some
value of n, say n=k, then it will always work for the
next integer n=k+1.
Therefore
Since we have shown it works for n=k=1, it works for n=k+1=2
Since it works for n=k+1=2, it works for n=k+2=3
Since it works for n=k+2=3, it works for n=k+3=4
Since it works for n=k+3=4, it works for n=k+4=5
etc. etc. forever
Edwin
So we assume1+5+9+....(4k-3) = k(2k-1).What we want to show is that if we substitute k+1 for n in1+5+9+...(4n-3) = n(2n-1)It will also work.That is we want to show that based on the assumption that it works for k, that we'll end up with this without the questionmark over the equal sign:1+5+9+...+[4(k+1)-3] ≟ (k+1)[2(k+1)-1] The question mark above the equal sign is to show that wehave not shown that yet. The above is what we must show.We will simplify what we are to show:1+5+9+...+[4k+4-3] ≟ (k+1)[2k+2-1]1+5+9+...+(4k+1) ≟ (k+1)(2k+1)1+5+9+...+(4k+1) ≟ 2k²+3k+1Now we will show it. We start with:1+5+9+...+(4k-3) = k(2k-1)We add the k+1st term, which is (4k+1), to both sides,and hope that the right side comes out to to theexpression above, which is 2k²+3k+1:1+5+9+...+(4k-3)+(4k+1) = k(2k-1)+(4k+1)1+5+9+...+(4k-3)+(4k+1) = 2k²-k+4k+1 1+5+9+...+(4k-3)+(4k+1) = 2k²+3k+1So we have shown that if the formula works for somevalue of n, say n=k, then it will always work for thenext integer n=k+1.ThereforeSince we have shown it works for n=k=1, it works for n=k+1=2Since it works for n=k+1=2, it works for n=k+2=3Since it works for n=k+2=3, it works for n=k+3=4Since it works for n=k+3=4, it works for n=k+4=5etc. etc. foreverEdwin
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