Evaluate the sum using the summation formulas and properties.

Question

12

(j³ - 3j²) j = 1

= ______

Answer 1

(1^3 - 3(1)^2 = 1-3 =-2

2^3 -3(2)^2 = 8-12 = -4

3^3 -3(3)^2 = 0

4^3 -3(4)^2 = 64-48 = 16

5^3 -3(5^2 = 125-75 = 50

6^3 -3(6^2 = 216 -108 = 108

7^3 -3(7^2 = 196

8^3 -3(8)^2 = 512 - 192 = 320

9^3 -3(9)^2 = 729 - 243 = 486

10^3 -3(10^2 = 1000-300= 700

11^3 -3(11)^2 = 1331 - 363 = 968

12^3 -3(12)^2 = 9(144) = 1296

add them up, it's a few thousand

sum of j^3 - 3j^2 from j=1 to j=12 is 4134

which shows the formula used by Yefim works

Answer 2

∑₁¹²(j³ - 3j²) = [12(12 + 1)/2]² - 3·12·(12 + 1)(2·12 + 1)/6 = (6·13)² - 3·2·13·25 = 6084 - 1950 = 4134