✔ 最佳答案
n=1時,
左式 = (2*1-1)^2 - (2*1)^2 = 1 - 4 = - 3
右式 = - 1(2*1+1) = - 3
故 n = 1 時成立.
設 n = k 時該式成立,即:
1^2 - 2^2 + 3^2 - 4^2 + ..... + (2k-1)^2 - (2k)^2 = -k(2k+1) .....(1式)
當 n = k+1 時:
左式
= 1^2 - 2^2 + 3^2 - 4^2 + ..... + (2k-1)^2 - (2k)^2 + (2(k+1)-1)^2 - (2(k+1))^2
= -k(2k+1) + (2(k+1)-1)^2 - (2(k+1))^2 (由1式)
= -k(2k+1) + (2k+1)^2 - (2k+2)^2
= -k(2k+1) + (2k+1+2k+2)(2k+1-2k-2) (利用公式a^2-b^2 = (a+b)(a-b) )
= -k(2k+1) + (4k+3)(-1)
= -2k^2 -k -4k -3
= -2k^2 -5k -3
= -(2k^2+5k+3)
= -(k+1)(2k+3)
= -(k+1)(2(k+1)+1)
= 右式
所以 n = k+1 時該式亦成立
故得證