1) Prove that 1 + 4 + 7+....+ (3n-2) = (3n^2 - n)/2 .
2) Use induction to show that n^2 - n is always even.
Can someone help me with these problems? I have to explain this to my group so that we can present. Thanks a lot!!!
Well if you know $\displaystyle \sum_{i=1}^{n}{x_{i}} \frac{n(n+1)}{2}$ you can avoid induction for the first part, and write it as.
$\displaystyle \sum_{i=1}^{n}{(3x_{i} -2)} = 3\sum_{i=1}^{n}{x_{i}} - 2n$
Or if you prefer induction.
assume $\displaystyle \sum_{i=1}^{k}{(3x_{i} -2)} = \frac{3n^2-n}{2}$
then $\displaystyle \sum_{i=1}^{k+1}{(3x_{i} -2)} = \sum_{i=1}^{k}{(3x_{i} -2)} +3(k+1) - 2$
and i am sure you can finish that off.
the second part can be factored into the product of two consecutive numbers, so it should be fairly obvious that it must be even.
however you have been asked to do it by induction. so
suppose $\displaystyle n(n-1)$ is even for all n up a value k
then $\displaystyle k(k-1)$ is even. now we need to show that if $\displaystyle n = k+1$ it is still even
replace n with k+1 giving $\displaystyle (k+1)(k)$ expand it $\displaystyle k^2 + k$ then you add "zero" (i.e add k and then subtract k )
giving $\displaystyle k^2 +k +k - k \Rightarrow k^2 - k + 2k$
$\displaystyle \Rightarrow k(k-1) + 2k$
which is even as $\displaystyle k(k-1)$ is even
Hello, Leilei!
Verify $\displaystyle S(1)\!:\;\;1 \:=\:\frac{3\!\cdot\!1^2-1}{2} \:=\:1$ . . . True!1) Prove that:.$\displaystyle 1 + 4 + 7+ \hdots + (3n-2) \:= \:\frac{3n^2 - n)}{2}$
Assume $\displaystyle S(k)\!:\;\;1 + 4 + 7 + \hdots + (3k-2) \;=\;\frac{3k^2-k}{2}$
Add $\displaystyle 3(k+1)-2 \:=\:3k+1$ to both sides:
. . $\displaystyle 1 + 4 + 7 + \hdots + [3k-2] + [3(k+1)-2] \;=\;\frac{3k^2 - k}{2} + 3k+1 $
The left side is the left side of $\displaystyle S(k+1).$
The right side is: . $\displaystyle \frac{3k^2-k}{2} + 3k+1 \;=\;\frac{3k^2-k + 6k+2}{2} \;=\;\frac{3k^2 + 5x + 2}{2}$
. .. . $\displaystyle = \;\frac{(k+1)(3k+2)}{2} \;=\;\frac{(k+1)(3[k+1]-1)}{2} \;=\;\frac{(k+1)^2 - 3(k+1)}{2}$
This is the right side of $\displaystyle S(k+1)\;\;\hdots$ . The inductive proof is complete.
$\displaystyle S(n)\!:\;\;n^2-n \:=\:2a\;\text{ for some integer }a.$2) Use induction to show that $\displaystyle n^2 - n$ is always even.
Verify $\displaystyle S(1)\!:\;\;1^2-1 \:=\:0 \:=\:2(0)$ . . . True!
Assume $\displaystyle S(k)\!:\;\;k^2-k \:=\: 2b\;\text{ for some integer }b.$
Add $\displaystyle 2k$ to both sides: .$\displaystyle k^2 + k \;=\;2b+2k$
The right side is: .$\displaystyle 2(b + k)$ ... an even number.
Add and subtract $\displaystyle 2k+1$ to the left side:
. . $\displaystyle k^2 \:{\color{blue}+\: 2k + 1} + k \:{\color{red}- \:2k - 1} \;=\;(k^2+2k+1) - k - 1 \;=\;(k+1)^2-(k+1)$
And we have: .$\displaystyle (k+1)^2-(k+1) \;=\;2(b+k) $
We have proved $\displaystyle S(k+1)\;\;\hdots$ .The proof is complete.