proofs
- Thread starter karliekay
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Hello, karliekay!
\(\displaystyle \text{The sum of an arithmetic series is: }\:S \;=\;\tfrac{n}{2}[2a + (n-1)d]\)
. . \(\displaystyle \text{where: }\:\begin{Bmatrix} n &=& \text{number of terms} \\ a &=& \text{first term} \\ d &=& \text{common difference} \end{Bmatrix}\)
\(\displaystyle \text{Our series has: }\:a = 1,\;d = 2,\:\text{ and }n\text{ terms.}\)
\(\displaystyle \text{Therefore: }\:S \;=\;\tfrac{n}{2}[2(1) + (n-1)2] \;=\;\tfrac{n}{2}[2 + 2n - 2] \;=\;\tfrac{n}{2}[2n] \;=\;n^2\)
\(\displaystyle \text{The sum of an arithmetic series is: }\:S \;=\;\tfrac{n}{2}[2a + (n-1)d]\)
. . \(\displaystyle \text{where: }\:\begin{Bmatrix} n &=& \text{number of terms} \\ a &=& \text{first term} \\ d &=& \text{common difference} \end{Bmatrix}\)
\(\displaystyle \text{Our series has: }\:a = 1,\;d = 2,\:\text{ and }n\text{ terms.}\)
\(\displaystyle \text{Therefore: }\:S \;=\;\tfrac{n}{2}[2(1) + (n-1)2] \;=\;\tfrac{n}{2}[2 + 2n - 2] \;=\;\tfrac{n}{2}[2n] \;=\;n^2\)
How about a visual demonstration?
. . \(\displaystyle \square\)
\(\displaystyle 1 \:=\:1^2\)
. . .\(\displaystyle \begin{array}{c} \square \blacksquare \\ [-1mm] \blacksquare \blacksquare \end{array}\)
\(\displaystyle 1 + 3 \:=\:2^2\)
. . . \(\displaystyle \begin{array}{c}\square\square\blacksquare \\ [-1mm]\square\square\blacksquare \\ [-1mm] \blacksquare\blacksquare\blacksquare \end{array}\)
\(\displaystyle 1 + 3 + 5 \:=\:3^2\)
. . . .\(\displaystyle \begin{array}{c}\square\square\square\blacksquare \\[-1mm] \square\square\square\blacksquare \\[-1mm] \square\square\square\blacksquare \\[-1mm] \blacksquare\blacksquare\blacksquare\blacksquare \end{array}\)
\(\displaystyle 1 + 3 + 5 + 7 \:=\:4^2\)
. . . . \(\displaystyle \begin{array}{c}\square\square\square\square\blacksquare \\[-1mm] \square\square\square\square\blacksquare \\[-1mm] \square\square\square\square \blacksquare \\[-1mm] \square\square\square\square\blacksquare \\[-1mm]\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare \end{array}\)
\(\displaystyle 1 + 3 + 5 + 7 + 9 \:=\:5^2\)
karliekay said:
How about a proof by induction?
Is the given statement true when n = 1?
1 = 1[sup:2jy35jtt]2[/sup:2jy35jtt]
1 = 1
True
Assume that the given statement is true when n = k...that is, assume that
1 + 3 + 5 + 7 + ... + (2k - 1) = k[sup:2jy35jtt]2[/sup:2jy35jtt]
Will it be true for k + 1?
The term AFTER (2k - 1) would be (2k - 1) + 2, or (2k + 1). Let's add (2k + 1) to both sides of the equation:
1 + 3 + 5 + 7 + .... + (2k - 1) + (2k - 1 + 2) = k[sup:2jy35jtt]2[/sup:2jy35jtt] + (2k + 1)
1 + 3 + 5 + 7 + .....+ (2k - 1) + (2k + 1) = k[sup:2jy35jtt]2[/sup:2jy35jtt] + 2k + 1
but k[sup:2jy35jtt]2[/sup:2jy35jtt] + 2k + 1 is (k + 1)[sup:2jy35jtt]2[/sup:2jy35jtt]
1 + 3 + 5 + 7 + .....+ (2k - 1) + (2k + 1) = (k + 1)[sup:2jy35jtt]2[/sup:2jy35jtt]
mmm4444bot
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soroban said:
How about a visual demonstration?
It's a lovely demonstration (on more than one level), and I like it, but I do not believe that it constitutes a proof.