2

Standard proof that $\sum_1^n i^2=\frac{n(n+1)(2n+1)}{6}$ is given by induction. It is of course a valid method. However, I find that it requires me to memorize the answer. Is there any way I can sit in an exam and just derive the result from scratch without memorization?

This question is apparently more general and different than Gaussian proof for the sum of squares? and I have seen quite a few great answers other than the "Gaussian method" in the post

Daniel Li
  • 3,200
  • You could write $i^2$ in binomial notation as $2\binom i2+2\binom i1$ and use binomial identities... – abiessu Aug 14 '19 at 17:50
  • 4
    One way is to first start by deriving the value of the (telescoping) sum: $$\sum_{i=1}^n (i+1)^3-i^3$$ Then by expanding the summand and by the linearity property, infer the value of $\sum_{i=1}^n i^2$ by using the (well-known) values of $\sum_{i=1}^n i$ and $\sum_{i=1}^n 1$. – projectilemotion Aug 14 '19 at 17:52
  • https://en.wikipedia.org/wiki/Hockey-stick_identity – Jean-Claude Arbaut Aug 14 '19 at 17:52
  • 1
    Try guessing that it will be a cubic polynomial and solve for the coefficients. – Jair Taylor Aug 14 '19 at 17:53
  • This older question is very much in the vein of the present one: https://math.stackexchange.com/q/2979513/137524 – Semiclassical Aug 14 '19 at 17:55
  • Some time ago, I proved it by breaking up the summation such that we can use the known formula for $\sum_1^n i$. – JG123 Aug 14 '19 at 17:56
  • 1
    The simpler sums can be written as $$\binom{m+1}{k+1}=\sum_{n=0}^{m}\binom{n}{k}$$ which can be proved combinatorially. Then $n^2=2\binom{n}{2}+\binom{n}{1},$ so $$\begin{align}\sum_{n=0}^{m} n^2 &= 2\binom{m+1}{3}+\binom{m+1}{2}\&=\frac{(m+1)m(m-1)}{3}+\frac{(m+1)m}{2}\&=\frac{m(m+1)}{6}\left(2(m-1)+3\right)\&=\frac{m(m+1)(2m+1)}{6}\end{align}$$ – Thomas Andrews Aug 14 '19 at 18:09
  • The reason that $$\binom{m+1}{k+1}=\sum_{n=0}^{m}\binom{n}{k}$$ is that $\binom{m+1}{k+1}$ counts the number of subsets of size $k+1$ of the set ${0,1,\dots,m}$ and $\binom{n}{k}$ counts those subsets with maximum element $n.$ – Thomas Andrews Aug 14 '19 at 18:19
  • So, for example, we have $$n^3=6\binom{n}{3}+6\binom{n}{2}+\binom{n}{1}$$ and you get $$\sum_{n=0}^{m} m^3 = 6\binom{n+1}{4}+6\binom{n+1}{3}+\binom{n+1}{2}$$ – Thomas Andrews Aug 14 '19 at 18:27
  • 3
    Another approach to discover the formula is to use a difference table, where the first row contains the first several values of the sequence, starting with $n=0$, and each subsequent row contains the differences of adjacent values in the row above it: \begin{align} 0 && 1 && 5 && 14 && 30 \ 1 && 4 && 9 && 16 \ 3 && 5 && 7 \ 2 && 2 \ 0 \end{align} Now read off the first column as multipliers for binomial coefficients: $$\sum_{i=1}^n i^2 = 0 \binom{n}{0}+1 \binom{n}{1}+3 \binom{n}{2}+2 \binom{n}{3}=\frac{n (n + 1) (2 n + 1)}{6}$$ – RobPratt Aug 14 '19 at 19:44
  • See also https://math.stackexchange.com/questions/48080/sum-of-first-n-squares-equals-fracnn12n16 – leonbloy Aug 15 '19 at 12:57

0 Answers0