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所有问题
1x2+2x3+3x4+...+n(n+1)=? 1x2x3+2x3x4+3x4x5+...+n(n+1)(n+2)=?
如题所述
举报该问题
推荐答案 2012-02-11
(1*1+1)+(2*2+2).......=(1*1+2*2+3*3......+N*N)+(1+2+3+4+...+N)=
同理,第二题乘出来:(1*1*1+2*2*2+3*3*3+......n*n*n)+3(1*1+2*2...+n*n)+2(1+2+3+4+..+n)=
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其他回答
第1个回答 2012-02-11
n(n+1)(n+2)/3
n(n+1)(n+2)(n+3)/4
......
定义:
n(n+1)(n+2)...(n+k)=[n]^k
则:
∑(i=1 to n)[n]^k=[n]^(k+1)/(k+1)=n(n+1)...(n+k+1)/(k+1)
追问
看不懂啊,能不能直接给答案。
追答
开始2行就是答案啊。第一行是第一题答案,第二行是第二题答案。
本回答被提问者采纳
第2个回答 2012-02-11
数列求和
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n(n+1)=
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...
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1x2+2x3+3x4+
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1/4[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)+(n-1)n(n+1)(n+2)...-0*1*2*3)]=1...
1x2
2x3
3x4
...
n(n
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答:
1x2+2x3+3x4+
...
+n(n+1)=
1^2+1+2^2+2+3^2+3+...+n^2+n =1
+2+
...+n+(1^2+2^2+...+n^
2)=
(1+n)n/2
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2n+1)/6 =n(n+1)/2*(1+(2n+1)/3)=n(n+1)(2n+5)/6
1x2+2x3+3x4+4x5+.+n(n+1)
等于多少?急救! 请写下过程,谢谢
答:
1x2+2x3+3x4+
…
+n(n+1) =
1^2+1+2^2+2+3^2+3+…+n^2+n =(1^2+2^2+3^2+…+n^2)+(1+2+3+…+n) =1/6*n(n+1)(2n+1)+1/2*n(n+1) =1/6*n(n+1)(2n+1+3)(提取公因式) =1/3*
n(n+1)(n+2)
1x2+2x3+3x4+4x5+.+n(n+1)
等于多少?急救! 请写下过程,谢谢
答:
1x2+2x3+3x4+
…
+n(n+1) =
1^2+1+2^2+2+3^2+3+…+n^2+n =(1^2+2^2+3^2+…+n^2)+(1+2+3+…+n) =1/6*n(n+1)(2n+1)+1/2*n(n+1) =1/6*n(n+1)(2n+1+3)(提取公因式) =1/3*
n(n+1)(n+2)
...不然我看不懂 还有解释下
1X2+2X3+3X4+
...
+n+(N
加
1)=
答:
1x2+2x3+3x4+
…
+n(n+1)=1x
(1+1)+2x(2+1)+3x(3+1)+…n(n+1)=(1^2+2^2+3^2+…+n^2)+(1+2+3+…+n)=n(n+1)(2n+1)/6+n(n+1)/2 =n(n+1)[(2n+1)+3]/6 (1) (1+2+3+…+n) =n(n+1)/2 (2) (1^2+2^2+3^2+…+n^
2)=n(n+1)(
2n+1...
求解,Sn=
1x2+2x3+3x4+
…
+n(n+1)=?
跪求,要过程
答:
裂项相消法。原式:
1x2+2x3+3x4+
...
+n(n+1)=
1^2+1+2^2+2+3^2+3+...+n^2+n =1
+2+
...+n+(1^2+2^2+...+n^
2)=
(1+n)n/2
+n(n+1)(
2n+1)/6 =n(n+1)/2*(1+(2n+1)/3)=n(n+1)(2n+5)/6 欢迎采纳!我帮你!!
1x2+2x3+3x4+
...
+ n
的计算方法
答:
求
1x2+2x3+3x4+
……
+n(n+1)
注意到:(n+1)^3-n^3=3n^2+3n+1 则可得:
n(n+1)=
[(n+1)^3-n^3]/3-1/3 那么有:1×2=(2^3-1^3)/3-1/3 2×3=(3^3-2^3)/3-1/3 ……累加可得:所求算式 =[(n+1)^3-1^3]/3-n/3 =(n^3+3n^2+3n-n)/3 =n(n+1)...
1x2+2x3+3x4+4x5+
...
+n(n+1)
等于多少?急救!!
答:
1^
2+2
^
2+3
^2+……+n^2
=n(n+1)(
2n+1)/6 利用立方差公式 n^3-(n-1)^
3=1
*[n^2+(n-1)^
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^2-2 3^3-2^3=2*3^2+2^2-3 4^3-3^3=2*4^2+3^2-4 ...n^3-(n-1)^3=2...
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