Stuck on proof by induction

Hello,

I start to solve this, but got stacked. Have no idea how to move on :confused:

Anybody help me ? (Thinking)

Code:

`5+8+11+…+ (3n+2) = 1/2n(3n+7)`

n=1

5+8+11+…+ (3*1+2) = 1/2 *1 (3*1+7) => 5=5

n=k

5+8+11+…+ (3k+2) = 1/2 k(3k+7)

n=k+1

5+8+11+…+ (3(k+1)+2) = 1/2(k+1) (3(k+1)+7)

5+8+11+…+ (3k+5) = 1/2 (k+1) (3k+10)

5+8+11+…+ (3k+2) + (3k+5) = 1/2 (k+1) (3k+10)

1/2 k(3k+7) + (3k+5) = 1/2 (k+1) (3k+10)

1/2 [3k^2 + 10k + 5] = 1/2 (k+1) (3k+10)

=>

Re: Stuck on proof by induction

Hello everybody, again me :D

Now i'm trying to solve or prove this one:

1^2+2^2+3^2+....+n^2=n(n+1)(2n+1)/6

And I came to the third step where i got this one:

2k^3+10k^2+13k+6 on the LHS. I know that to solve this and get the same like on RHS, I need to work with division with polynoms, but even if i try to do that, still can't prove LHS=RHS.

Can anybody help me somehow? (Worried)

Thanks

Re: Stuck on proof by induction

I think it should be (2k^3+**9**k^2+13k+6) / 6.

Re: Stuck on proof by induction

One mistake and everything is screwed up :D But it's great when there is someone that can fix that :) Thanks a lot, I solve it

Re: Stuck on proof by induction

Quote:

Originally Posted by

**Scorpy6** Hello everybody, again me :D

Now i'm trying to solve or prove this one:

1^2+2^2+3^2+....+n^2=n(n+1)(2n+1)/6

And I came to the third step where i got this one:

2k^3+10k^2+13k+6 on the LHS. I know that to solve this and get the same like on RHS, I need to work with division with polynoms, but even if i try to do that, still can't prove LHS=RHS.

Can anybody help me somehow? (Worried)

Thanks

Base Step: $\displaystyle \displaystyle n = 1$

$\displaystyle \displaystyle \begin{align*} LHS &= 1^2 \\ \\ RHS &= \frac{1(1 + 1)(2\cdot 1 + 1)}{6} \\ &= \frac{1 \cdot 2 \cdot 3}{6} \\ &= 1 \\ &= LHS \end{align*} $

Inductive Step: Assume the statement is true for $\displaystyle \displaystyle n = k$, so assume $\displaystyle \displaystyle 1^2 + 2^2 + 3^2 + \dots + k^2 = \frac{k(k+1)(2k+1)}{6} $.

Now we need to prove the statement is true for $\displaystyle \displaystyle n = k + 1$, i.e. show that $\displaystyle \displaystyle 1^2 + 2^2 + 3^2 + \dots + k^2 + (k + 1)^2 = \frac{(k+1)(k + 2)(2k + 3)}{6}$.

$\displaystyle \displaystyle \begin{align*} LHS &= 1^2 + 2^2 + 3^2 + \dots + k^2 + (k + 1)^2 \\ &= \frac{k(k + 1)(2k + 1)}{6} + (k + 1)^2 \\ &= \frac{k(k + 1)(2k + 1) + 6(k + 1)^2}{6} \\ &= \frac{(k + 1)\left[k(2k + 1) + 6(k + 1)\right]}{6} \\ &= \frac{(k + 1)(2k^2 + k + 6k + 6)}{6} \\ &= \frac{(k + 1)(2k^2 + 7k + 6)}{6} \\ &= \frac{(k+1)(2k^2 + 4k + 3k + 6)}{6} \\ &= \frac{(k + 1)[2k(k + 2) + 3(k + 2)]}{6} \\ &= \frac{(k + 1)(k + 2)(2k + 3)}{6} \\ &= RHS \end{align*}$

Q.E.D. :)

Re: Stuck on proof by induction

Or...

Show that

$\displaystyle 1^2+2^2+....+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6 }$

if

$\displaystyle 1^2+2^2+...+k^2=\frac{k(k+1)(2k+1)}{6}$

Then

$\displaystyle \frac{k(k+1)(2k+1)}{6}+\frac{6(k+1)^2}{6}=\frac{(k +1)(k+2)(2k+3)}{6}\;\;\;?$

$\displaystyle [k+1][k(2k+1)+6(k+1)]=[k+1][(2k+3)(k+2)]\;\;\;?$

$\displaystyle k(2k+1)+6(k+1)=(2k+3)(k+2)\;\;\;?$

$\displaystyle 2k^2+k+6k+6=2k^2+4k+3k+6\;\;\;?$