Work out the values of a and b (in the below question):

$$1^2 = 1$$

$$1^2 + 2^2 = 5$$

$$1^2 + 2^2 + 3^2 = 14$$

$$1^2 + 2^2 + 3^2 + 4^2 = 30$$

$$1^2 + 2^2 + 3^2 + 4^2 + ...... + n^2 = an^3 + bn^2 + (n/6)$$

Work out the values of $$a$$ and $$b$$.

$$\sum_{r=0}^nr^2=\frac{n(n+1)(2n+1)}{6}$$ now you can expand this

• just a question. – user547075 Nov 13 '18 at 3:00
• shouldn't r start with 1? – user547075 Nov 13 '18 at 3:00
• Possibly yes, although in this context the $r=0$ term is ignored obviously – Henry Lee Nov 13 '18 at 3:04

Sum of squares of first $$n$$ natural numbers is given by : $$\Sigma_{r=1}^nr^2= \frac{n(n+1)(2n+1)}{6}$$ Expanding it: $$\frac13 n^3+\frac 12 n^2+\frac n6$$

So, $$a=\frac 13$$ and $$b=\frac 12$$

• Thanks for the reply, but is it possible to get the values of a and b without using sigma? – V11 Nov 13 '18 at 4:40
• We used sigma to get that expression. Once got that, then just simplify it to form a cubic. And you are done! – idea Nov 13 '18 at 5:27

By one of the properties of sigma, the last equation can be expressed as $$\frac{n(n+1)(2n+1)}{6}$$.

This means that the sum of the squares of numbers from $$1$$ to $$n$$ can be expressed as $$\frac{n(n+1)(2n+1)}{6}$$.

Therefore, you just have to make it equal to that and work out the values of $$a$$ and $$b$$ by yourself.

To be more clear, you just need to let $$an^3+bn^2+\frac{n}{6}= \frac{n(n+1)(2n+1)}{6}$$ and solve for $$n$$.

• What do you mean exactly? – V11 Nov 13 '18 at 3:42
• I edited my answer – user547075 Nov 13 '18 at 4:20
• Thanks for the reply. I still have a few doubts though. How will solving for n get the values for a and b? And is it possible to get the values for a and b without using sigma? – V11 Nov 13 '18 at 4:44
• 1^2+2^2+.....+n^2=an^3+bn^2+n/6=n(n+1)(2n+1)6 Is it better to understand?In case you are curious about how the formula is made, you can copy and go to this link. mathforum.org/library/drmath/view/56920.html . This shows you how the formula is achieved. – user547075 Nov 13 '18 at 21:57