用归纳法证明:(1×2²-2×3²)+(3×4²—4×5²)+.+[(2n-1)
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/13 17:03:29
用归纳法证明:(1×2²-2×3²)+(3×4²—4×5²)+.+[(2n-1)(2n)²-2n(2n+1)²]=-n×(n+1)×(4n+3)
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
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
1²-2²+3²-4²+5²-6²+…-100²+
(2²+4²+6²+.+98²+100²)-(1²+3&su
1.计算:1²+4²+6²+7²=102,2²+3²+5&s
不用计算器求值1²+2²+3²+4²+5²+6²+7&sup
用数学归纳法证明:1²+2²+...+n²=n(n+1)(2n+1)/6 (n是正整数)
4²+3²>2*4*3,(-2)²+1²>2*(-2)*1,2²+2&
100²-99²+98²-97²+.4²-3²+2²
计算:100²-99²+98²-97²+…4²-3²+2&s
计算:100²-99²+98²-97²+……+4²-3²+2
已知:多项式(2mx²-x²+3x-1)-(5x²-4y²+3x²)不
(1-1/2²)(1-1/3²)(1-1/4²).(1-1/10²)=?
1.因式分解(1) 1-4x²y²(2) 16m²+25n²+40mn(3) 2