Question

用归纳法证明：（1×2²-2×3²）+（3×4²—4×5²）+.+[(2n-1)

Answer 1

n = 1 时

左端 = 1×2²-2×3² = -14

右端 = -1×(1+1)×(4×1+3) = -14

命题成立

设 n = k 时成立

（1×2²-2×3²）+（3×4²—4×5²）+.+[(2k-1)(2k)²-2k(2k+1)²]=-k×（k+1）×（4k+3）

n=k+1 时

左端 =

（1×2²-2×3²）+（3×4²—4×5²）+.+ [(2k-1)(2k)²-2k(2k+1)²] + [(2k+1)(2k+2)²-2（k+1)(2k+3)²]

= -k×（k+1）×（4k+3）+ [(2k+1)(2k+2)²-2(k+1)(2k+3)²]

= -k×（k+1）×（4k+3）+ 2(k+1)[2(2k+1)(k+1)-(2k+3)²]

= -k×（k+1）×（4k+3）+ 2(k+1)[4k² +6k + 2 -4k²-12k - 9]

= -k×（k+1）×（4k+3）- 2(k+1)(6k + 7)

= -(k+1)×[k×(4k+3) + 2(6k+7)]

= -(k+1)×[4k²+15k+14)

= -(k+1)×(k+2)×(4k+7)

右端 = -(k+1)×[(k+1）+1]×[4(k+1)+3]

= -(k+1)×(k+2)×(4k+7)

左端 = 右端

所以 n = k+1 时 命题成立

因此 原命题成立