(1)an=2n-1(2)465【分析】(1)根据等差数列的通项公式和求和公式列式求出a1和d,可得通项公式;(2)先求出Sn,再利用并项求和法与等差数列的求和公式可得结果.【详解】(1)设公差为d,则\((array)la_1+d+a_1+2d=8 5a_1+10d=25(array).,解得a1=1,d=2,所以an=1+(n-1)⋅2=2n-1.(...
am+an+bm+bn x^2+5y-xy-5x x2+xy+2xz+x (a+b)x-(b+a)y -12x(x-y)-(y-x)x^2 2x^2+4x+2 x(x+z)-y(y+z) 2(x^2-3ab)+x(4a-3b) 4x^2+12xy+9y^2-25 x^4-16 16x^4-1 x^2+9x+8 x^2-10xy+24y^2 2x^2+15xy-8y^2 3x^2-6xy+3y^2 24x^2+22x-21 -x^2...
记Sn为等差数列{an}的前n项和,已知a2+a3=8,S5=25. (1)求{an}的通项公式; (2)数列{bn}满足b1=1,bn+1=an-bn(n∈N*),求数列{bn}的前21项和. 【考点】数列的求和;等差数列的前n项和. 【答案】见试题解答内容 【解答】 【点评】
所以ak-ak+1是数列{an}中的项,是第 3k2+7k+2 2项. (Ⅲ)证明:由(II)知: an= 2 3n+2, bn= 2 3( 1 an+5)= 2 3( 3n+2 2+5)=n+4.下面用数学归纳法证明:2n+4>(n+4)2对任意n∈N*都成立.(1)当n=1时,显然25>52,不等式成立.(2)假设当n=k(k∈N*)时,有2k+4>(k+4)2,当n...
的图象上,且数列{an}是a1=1,公差为d的等差数列.(1)证明:数列{bn} 是公比为 ( 1 2)d的等比数列;(2)若公差d=1,以点Pn的横、纵坐标为边长的矩形面积为cn,求最小的实数t,若使cn≤t(t∈R,t≠0)对一切正整数n恒成立;(3)对(2)中的数列{an},对每个正整数k,在ak与ak+1之间插入2k-1个3(如在...
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解答解:数列{an}是等比数列,其公比为2, 设bn=log2an,且数列{bn}的前10项的和为25, ∴b1+b2+…+b10=log2(a1•a2•…•a10)=log2(a10121+2+…+9)log2(a11021+2+…+9)=25, ∴a101×245a110×245=225,可得:a1=1414. 那么1a11a1+1a21a2+1a31a3+…+1a101a10=4(1+12+122+…+129)...