The proof of k2-k-1 when k is greater than 2 The square of k-k-1 is always greater than 0 when k is greater than 2

The proof of k2-k-1 when k is greater than 2 The square of k-k-1 is always greater than 0 when k is greater than 2

k^2-k-1=k(k-1)-1＞2*1-1=1＞0

K1 = 2-k2 into K1 / 2 + 4k2 = 9 / 2, can you write details

k1/2 +4k2=9/2
Multiply both sides by two
k1 +8k2=9
Take K1 = 2-k2 into the above equation
2-k2+8k2=9
2+7k2=9
7k2=7
k2=1

Sequence sum: sequence Sn = 1 + 4 + 9 + +N ^ 2. Find SN
Hope to give a detailed proof process, is to evaluate, not to use mathematical induction to prove the answer

Method 1: because (n + 1) ^ 3 = n ^ 3 + 3N ^ 2 + 3N + 1, 2 ^ 3 = 1 ^ 3 + 3 * 1 ^ 2 + 3 * 1 + 13 ^ 3 = 2 ^ 3 + 3 * 2 + 3 * 2 + 14 ^ 3 = 3 ^ 3 + 3 * 3 ^ 2 + 3 * 3 + 15 ^ 3 = 4 ^ 3 + 3 * 4 ^ 2 + 3 * 4 + 1 n ^3 = (n-1)^3 + 3*(n-1)^2 + 3*(n...