(1/3)k(4k^2-1)+(2k+1)^2为什么=(1/3)(k+1)[4(k+1)^2-1]
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(1/3)k(4k^2-1)+(2k+1)^2为什么=(1/3)(k+1)[4(k+1)^2-1]
要什么公式?求详解
要什么公式?求详解
(1/3)k(4k^2-1)+(2k+1)^2
=(1/3)k(2k-1)(2k+1)+(2k+1)^2
=(1/3)(2k+1)[ k(2k-1) +3(2k+1) ]
=(1/3)(2k+1)(2k^2+5k+3)
=(1/3)(2k+1)(2k+3)(k+1)
=(1/3)(k+1)( 4k^2+8k+3 )
=(1/3)(k+1)( 4(k+1)^2 -1)
再问: ���һ���Ⱥ�û�п���Ŷ
再答: 4(k+1)^2 -1 =4k^2+8k +4 -1 =4k^2+8k +3
=(1/3)k(2k-1)(2k+1)+(2k+1)^2
=(1/3)(2k+1)[ k(2k-1) +3(2k+1) ]
=(1/3)(2k+1)(2k^2+5k+3)
=(1/3)(2k+1)(2k+3)(k+1)
=(1/3)(k+1)( 4k^2+8k+3 )
=(1/3)(k+1)( 4(k+1)^2 -1)
再问: ���һ���Ⱥ�û�п���Ŷ
再答: 4(k+1)^2 -1 =4k^2+8k +4 -1 =4k^2+8k +3
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