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1x2十2x3十3x4十公式
M=
1x2
+
2x3
+
3x4
+···+(n-1)xn,则M等于?
答:
M=
1
×
2
+2×
3
+...+(n-1)×n =[1×2×(3-0)+2×3×(4-1)+...+(n-1)×n×((n+1)-(n-2))]/3 =(1×2×3-0×1×2+2×3×4-1×2×3+...+(n-1)×n×(n+1)-(n-2)×(n-1)×n)/3...裂项 =(n-1)n(n+1)/3 ...消去中间项 ...
1x2十2x3十3x4十
···
十10x
11
答:
1×
2十2
×
3十3
×4十·.
十10
×11 =1²+1+2²+2+3²+3+.+10²+10 =(1²+2²+3²+.+10²)+(1+2+3+.+10)=10×(10+1)×(2×10+1)/6+10×(10+1)/2 =10×11×12/3 =440 ...
简算
1x2
+
2x3
+
3x4
+4x5+...+n(n+1)
答:
1x2
+
2x3
+
3x4
+4x5+...+n(n+1)=(1^2+2^2+……n^2)+(1+2+3+……n)=n(n+1)(2n+1)/6+(1+n)xn/2 =n(n+1)(n+2)/3
巧算
1x2
2x3
3x4
4x5…… 19x20
答:
1x2
+
2x3
+
3x4
+4x5+…… +19x20 =(1+3)×2+(3+5)×4+……+(17+19)×18+19×20 =2×4+4×8+6×12+……+18×36+19×20 =2×2×2+2×4×4+2×6×6+……+2×18×18+19×20 =2×(2²+4²+6²+……+18²)+19×20 =8×(1²+...
数学题
1x2
+
2x3
+
3x4
+。。。+2011x2012
答:
根据上面列出的规律,可以得出:(
1x2
+
2x3
+
3x4
+…+2011x2012)=(2011x2012x2013)/3 所以:1/(2011x2012)x(1x2+2x3+3x4+…+2011x2012)=2013/3
...3x4=
三
分之
一x
(
3x4x
5-
2x3x4
) 所以
1x2
+2x3+3
答:
(2)1x2x3=四分之
一x
(
1x2x3x4
-0x1x2x3),... , 7x8x9=四分之一x(7x8x9
x10
-6x7x8x9)故1x2x3+2x3x4+
3x4x
5+111+7x8x9=四分之一x(1x2x3x4-0x1x2x3)+... +四分之一x(7x8x9x10-6x7x8x9)= 四分之一x(7x8x9x10-0x1x2x3)=1260 (用结合率把1...
用裂项法求解(
1
*
2
+
3
*4+5*6+...+99*100)/(2*3+4*5+...+98*99)=?
答:
(1*2+3*4+5*6+...+99*100)/(1*2+2*3+4*5+...+98*99)=(99x100x101/3)/(98x99x100/3)=101/98
公式
推导:
1x2
+
2x3
+
3x4
+4x5+...+(n-1)xn =[1+2²+3²+...+(n-1)^2]+(1+2+3+...+n-1)=(n-1)n[2(n-1)+1]/6+n(n+1)/2 =(n-1)n[2...
1x2
+
2x3
+
3x4
+4x5+...+n(n+1)=?(n为正整数)
答:
回答:解 n(n+
1
)=n²+n ∴原式 =1+1²+
2
+2²+
3
+3²+……+n+n² =(1+2+3+……+n)+(1²+2²+3²+……+n²) =(1+n)n÷2+1/6n(n+1)(2n+1) =n(n+1)[1/2+1/6(2n+1)] =n(n+1)(1/3n+2/3) =1/3n...
1x2
+
2x3
+
3X4
…+99X100=?
答:
n(n+1)=(1/3) { n(n+1)(n+2) - (n-1)n(n+1) }
1x2
+2x3+3x4+...99x100 = 1x2 + (1/3) { (
2x3x4
- 1x2x3) + (
3x4x
5 - 2x3x4) +...+(99x100x101 - 98x99x100) } = 1x2 + (1/3) { 99x100x101 -1x2x3 } = (1/3) 99x100x101 =333300 ...
巧算
1x2
+
2x3
+
3x4
+4x5……+19x20 求过程+答案
答:
考察一般项:n(n+
1
)=n^2+n 1×
2
+2×
3
+...+19×20 =(1^2+2^2+...+19^2)+(1+2+...+19)=19×20×39/6 +19×20/2 =2470+190 =2660 一般的:1×2+2×3+...+n(n+1)=(1^2+2^2+...+n^2)+(1+2+...+n)=n(n+1)(2n+1)/6 +n(n+1)/2 =[n(n+1)...
<涓婁竴椤
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涓嬩竴椤
灏鹃〉
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