Question

solve with full induction:

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

Answer 1

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

Answer 2

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

Answer 3

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
\[\sum_{k=1}^{s+1}=1^2+2^2+\cdots + s^2 + (s+1)^2\]\[=\left[ (1/6)s(s+1)(2s+1) \right]+(s+1)^2\]

Answer 4

here i got stuck

Answer 5

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

Answer 6

\[1/3n^3+3/2n2+3/2n+1\]

Answer 7

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

Answer 8

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

Answer 9

thank you very much