giovkast 2 years ago solve with full induction: sum_{k=1}^{n} k^2 = 1/6n(n + 1)(2n+1)

1. giovkast

$\sum_{k=1}^{n} k^2 = 1/6n(n + 1)(2n+1)$

2. sirm3d

first step: n = 1: $\sum_{k=1}^{1}k^2=1^2 = 1,\quad (1/6)(1)(2)(3) = 1$

3. sirm3d

step 2: induction process. assume true for n=s $\sum_{k=1}^{s}=1^2+2^2+\cdots +s^2=(1/6) s(s+1)(2s+1)$ show the formula is true for n = s+1 $\sum_{k=1}^{s+1}=1^2+2^2+\cdots + s^2 + (s+1)^2$$=\left[ (1/6)s(s+1)(2s+1) \right]+(s+1)^2$

4. giovkast

here i got stuck

5. sirm3d

simplify the expressions, then factor. if you get $(1/6)(s+1)(s+2)(2s+3)$ then you're in the right track.

6. giovkast

$1/3n^3+3/2n2+3/2n+1$

7. giovkast

i got this, don't know where i went wrong

you need to add $\frac{s(s+1)(2s+1)}{6}+\frac{(s+1)^2}{1}=\frac{s(s+1)(2s+1)+6(s+1)^2}{6}$ $=\frac{(s+1)[s(2s+1)+6(s+1)]}{6}=\frac{(s+1)(2s^2+7s+6)}{6}=\frac{(s+1)(2s+3)(s+2)}{6}$ we need to show that this is the same as n=s+1 in the original ie $\frac{(\color{red}{s+1})(2\color{red}{(s+1)}+1)(\color{red}{s+1}+1)}{6}$TRUE

9. giovkast

thank you very much