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七年级数学解答题一般
题目
在数学中为了书写简便,1818世纪数学家欧拉就引进了求和符号"Σ\Sigma"(西格玛).
如记k=15k=1+2+3+4+5\sum_{k=1}^{5}{k}=1+2+3+4+5k=49k=4+5+6+7+8+9\sum_{k=4}^{9}{k}=4+5+6+7+8+9k=1nk=1+2+3++(n1)+n\sum_{k=1}^{n}{k}=1+2+3+⋯+(n-1)+nk=3n(x+k)=(x+3)+(x+4)++(x+n)\sum_{k=3}^{n}{(x+k)}=(x+3)+(x+4)+⋯+(x+n)
(1)(1)k=110k\sum_{k=1}^{10}{k}的值;
(2)(2)k=38(3x+k)\sum_{k=3}^{8}{(3x+k)}k=27(2x+k)\sum_{k=2}^{7}{(2x+k)}的差;
(3)(3)若对于任意xx都存在k=2n[x2+k(xa)]=4x2bx+20\sum_{k=2}^{n}{[{x}^{2}+k(x-a)]}=4{x}^{2}-bx+20,请分别求出aa,bb的值.
知识点:平方差公式、整式的混合运算章节:未标注

答案与解析

答案

(1)k=110k=1+2+3+4+5+6+7+8+9+10=55\sum_{k=1}^{10}{k}=1+2+3+4+5+6+7+8+9+10=55
(2)k=38(3x+k)=(3x+3)+(3x+4)+(3x+5)+(3x+6)+(3x+7)+(3x+8)=18x+33(2)\sum_{k=3}^{8}{(3x+k)}=(3x+3)+(3x+4)+(3x+5)+(3x+6)+(3x+7)+(3x+8)=18x+33
k=27(2x+k)=(2x+2)+(2x+3)+(2x+4)+(2x+5)+(2x+6)+(2x+7)=12x+27\sum_{k=2}^{7}{(2x+k)}=(2x+2)+(2x+3)+(2x+4)+(2x+5)+(2x+6)+(2x+7)=12x+27
k=38(3x+k)k=27(2x+k)=(18x+33)(12x+27)=6x+6\sum_{k=3}^{8}{(3x+k)}-\sum_{k=2}^{7}{(2x+k)}=(18x+33)-(12x+27)=6x+6
(3)k=2n[x2+k(xa)]=[x2+2(xa)]+[x2+3(xa)]++[x2+n(xa)](3)\sum_{k=2}^{n}{[{x}^{2}+k(x-a)]}=[{x}^{2}+2(x-a)]+[{x}^{2}+3(x-a)]+⋯+[{x}^{2}+n(x-a)]
=(n1)x2+(2+3++n)x(2a+3a++na)=\left(n-1\right)x^{2}+\left(2+3+\cdots +n\right)x-\left(2a+3a+\cdots +na\right)
k=2n[x2+k(xa)]=4x2bx+20\because \sum_{k=2}^{n}{[{x}^{2}+k(x-a)]}=4{x}^{2}-bx+20
(n1)x2+(2+3++n)x(2a+3a++na)=4x2bx+20\therefore \left(n-1\right)x^{2}+\left(2+3+\cdots +n\right)x-\left(2a+3a+\cdots +na\right)=4x^{2}-bx+20
n1=4\therefore n-1=4b=14b=-1414a=20-14a=20
a=107\therefore a=-\frac{10}{7}.

解析

(1)k=110k=1+2+3+4+5+6+7+8+9+10=55\sum_{k=1}^{10}{k}=1+2+3+4+5+6+7+8+9+10=55
(2)k=38(3x+k)=(3x+3)+(3x+4)+(3x+5)+(3x+6)+(3x+7)+(3x+8)=18x+33(2)\sum_{k=3}^{8}{(3x+k)}=(3x+3)+(3x+4)+(3x+5)+(3x+6)+(3x+7)+(3x+8)=18x+33
k=27(2x+k)=(2x+2)+(2x+3)+(2x+4)+(2x+5)+(2x+6)+(2x+7)=12x+27\sum_{k=2}^{7}{(2x+k)}=(2x+2)+(2x+3)+(2x+4)+(2x+5)+(2x+6)+(2x+7)=12x+27
k=38(3x+k)k=27(2x+k)=(18x+33)(12x+27)=6x+6\sum_{k=3}^{8}{(3x+k)}-\sum_{k=2}^{7}{(2x+k)}=(18x+33)-(12x+27)=6x+6
(3)k=2n[x2+k(xa)]=[x2+2(xa)]+[x2+3(xa)]++[x2+n(xa)](3)\sum_{k=2}^{n}{[{x}^{2}+k(x-a)]}=[{x}^{2}+2(x-a)]+[{x}^{2}+3(x-a)]+⋯+[{x}^{2}+n(x-a)]
=(n1)x2+(2+3++n)x(2a+3a++na)=\left(n-1\right)x^{2}+\left(2+3+\cdots +n\right)x-\left(2a+3a+\cdots +na\right)
k=2n[x2+k(xa)]=4x2bx+20\because \sum_{k=2}^{n}{[{x}^{2}+k(x-a)]}=4{x}^{2}-bx+20
(n1)x2+(2+3++n)x(2a+3a++na)=4x2bx+20\therefore \left(n-1\right)x^{2}+\left(2+3+\cdots +n\right)x-\left(2a+3a+\cdots +na\right)=4x^{2}-bx+20
n1=4\therefore n-1=4b=14b=-1414a=20-14a=20
a=107\therefore a=-\frac{10}{7}.

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