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1×2+2x3+3x4+4x5+n(n+1)
1x
2+2x3+3x4+4x5+
...
+n(n+1)
=?(n为正整数) 上面式子结果是多少?速...
答:
∴原式 =1+1²+2+2²+
3+3
²+……
+n+n
²=(
1+2+
3+……
+n)+
(1²+2²+3²+……+n²)=(
1+n)
n÷2+1/6
n(n+1)
(2n+1)=n(n+1)[1/
2+1
/6(2n+1)]=n(n+1)(1/3n+2/3)=1/3n(n+1)(n+2)公式 1²+2²+3...
1x
2+2x3+3x4+4x5+
...
+n(n+1)
=?(n为正整数)
答:
∴原式 =1+1²+2+2²+
3+3
²+……
+n+n
²=(
1+2+
3+……
+n)+
(1²+2²+3²+……+n²)=(
1+n)
n÷2+1/6
n(n+1)
(2n+1)=n(n+1)[1/
2+1
/6(2n+1)]=n(n+1)(1/3n+2/3)=1/3n(n+1)(n+2)公式 1²+2²+3...
1
x
2+2x3+3x4+4x5+
5x6+6x7+7x8+8x9=
答:
1x2 + 2x3 + 3x4 + .
+ n(n+1)
= 1/3 * n(n+1)(n+
2)
1x
2+2x3+3x4+4x5+
5x6+6x7+7x8+8x9 = 1/3 * 8 * 9 * 10 = 240
1x
2+2x3+3x4+4x5
……
+n(n+1)
(n+2)等于多少?
答:
1x
2+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)打字不易,如满意,望采纳。
1×2+2x3+3x4+4x5
简算方法是什么?
答:
根据公式可计算:
n(n+1)
=n^2
+n
1×2+2x3+3x4+4x5
=(1^2+2^2+...5^
2)
+(1+2+...+5)=(1+4+9+16+25)+(1+2+...+5)=55+15 =70 混合运算:如果一级运算和二级运算,同时有,先算二级运算。如果一级,二级,三级运算(即乘方、开方和对数运算)同时有,先算三级运算...
巧算
1
x
2+2x3+3x4+4x5
……+19x20
答:
考察一般项:n(n+1)=n^2+n1×2+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(...
1x2
2x3
3x4
...
n(n
1)
=?
答:
1x
2+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
1
x
2+2x3+3x4+4x5
...99x100
答:
设(1X2)/2+(2X3)/2+(3X4)/2+.………+
(n+1)
n/2=S 得1x
2+2x3+3x4+
………+(n+1)n=2S 由
n(n
-1)+(n+1)n=n^2-n
+n
^2+n=2n^2 当n为偶数时S=2^2+4^2+6^2+8^2+………+n^2 当n为奇数时S=2^2+4^2+6^2+8^2+………+(n-1)^2+(n+1)n/2 当n为偶...
计算
3
×
4+4×5+5×
6+...+12×13则么算
答:
+n(n+1)
=n(n+1)(n+2)/3;所以我们站在巨人的肩膀上得出:1x
2+2x3+3x4+4x5+
5x6+6x7+…+12x13=12(12+1)(12+2)/3;1x2+2x3+3x4+4x5+5x6+6x7+…+12x13=4(13)(14)=728;所以:
3×
4+4×5+
5×
6+...+12×13 =728-2x3-1x2 =728-6-2 =720 ...
1x
2x3+2x3x4+3x4x5+
...
+n(n+1)
(n+2)=?求此类型算数题公式!
答:
1×2
×
3+2×
3×
4+3×
4×
5+
...+n(n+1)(n+2)=(1³+2³+3³+...+n³)
+3(
1²+2²+3²+...+n²)+
2(
1+2+...+n)=[n(n+1)/2]²+n(n+1)(2n+1)/
2+n(n+1)
=n(n+1)[n(n+1)/4+(2n+1)/2+1]=n(n+1)...
<涓婁竴椤
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涓嬩竴椤
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