prove that [1^{2}+3^{2}+...+(2n-1)^{2}=(4n ^{3}-n)/3] for all natural number n

Question

turingtest · Answer

step 1) see if it is true for $n=1$

turingtest · Answer

$$1^2={4(1)^3-(1)\over3}$$$$\LARGE\checkmark$$step 2) assume it is true for some $n$...

turingtest · Answer

$$\sum_{i=0}^n(2i+1)^2={4n-n\over3}$$step 3) see if this implies that it is true for $n+1$

turingtest · Answer

sorry that should be$$1^2+3^2+...+(2n-1)^2={4n^3-n\over3}$$the next step is then proving this implies it is true for $n+1$$$1^2+3^2+...+(2n-1)^2+[2(n+1)-1]^2$$$$=1^2+3^2+...+(2n-1)^2+(2n+1)^2$$$$={4n^3-n\over3}+(2n+1)^2$$you will know that you have proven it if you can show that this last line equals$$4(n+1)^3-(n+1)\over3$$

anonymous · Answer

ok thanks

anonymous · Answer

did not work

turingtest · Answer

it does work...$$\frac{4n^3-n}3+(2n+1)^2$$$$\frac{4n^3-n}3+4n^2+4n+1$$$$\frac{4n^3+12n^2+12n-n+3}3$$$$\frac{4n^3+12n^2+12n+4-n-1}3$$$$\frac{4(n^3+3n^2+3n+1)-(n+1)}3$$$$\frac{4(n+1)^3-(n+1)}3~~~~\huge\checkmark$$