1x2+2x3+3x4公式推导
迟友坚19191x2+2x3+3x4+...+n(n+1)=?1x2x3+2x3x4+3x4x5+...+n(n+1)(n+2)=? -
宦通翁15720698322 ______[答案] 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)
迟友坚19191X2+2X3+3X4+.NX(N+1)规律 -
宦通翁15720698322 ______[答案] 这个数列的通项an = n(n+1)即 an = n^2 + nn(n+1)(2n+1)其中1^2 + 2^2 + 3^2 + ...+ n^2 = --------------6n(n+1)1+2+3+...+n = ------------2所以1X2+2X3+3X4+.NX(N+1) N(N+1)(2N+1) N(N+1)= ---------------- + --...
迟友坚19191x2+2x3+3x4+.+n(n+1) -
宦通翁15720698322 ______[答案] 先看n(n+1)=n2+n 把每一项都变成这样, 然后首先把所有二次方项加起来:1+2的平方+3的平方+...+n的平方,这个有公式可以算出来:n(n+1)(2n+1)/6 其次剩下的就是1+2+3+.+n 利用首尾相加法算出来:n(n+1)/2 两个结果的和就是最终答案.
迟友坚19191x2+2x3+3x4+4x5+.+n(n+1)等于多少?急救!请写下过程,谢谢 -
宦通翁15720698322 ______[答案] 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)
迟友坚19191x2+2x3+3x4+4x5+...+n(n+1)=?(n为正整数)上面式子结果是多少?速求. -
宦通翁15720698322 ______[答案] 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(n+1)(n+2) 公式 1²+2²+3²+……+n²=1/6n(n+1)(2n+1)
迟友坚19191x2+2x3+3x4+4x5+.+n(n+1)+n(n+2)=? -
宦通翁15720698322 ______[答案] 3n*(n+1) = n(n+1)(n+2) - (n-1)n(n+1) 所以3S = (n+1)(n+2)(n+3) - 0 S = (n+1)(n+2)(n+3) /3
迟友坚19191x2+2x3+3x4+…+29x30 -
宦通翁15720698322 ______[答案] 由1*2=(1/3)*1*2*3; 1*2+2*3=(1/3)*2*3*4; 1*2+2*3+3*4=(1/3)*3*4*5 可知 1*2+2*3+3*4+…+29*30 =(1/3)*29*30*31 =8990 同学你好,如果问题已解决,记得右上角采纳哦~~~您的采纳是对我的肯定~谢谢哦
迟友坚19191x2+2x3+3x4+...+n(n+1)=? -
宦通翁15720698322 ______[答案] 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---看通项,再分解,利用公式!
迟友坚19191x2+2x3+3x4+.+999x1000 -
宦通翁15720698322 ______[答案] 每一项这样拆:n*(n+1)=1/3[n*(n+1)(n+2)-(n-1)n(n+1)] 1x2+2x3+3x4+.+999x1000 =[(1x2x3-0x1x2)+(2x3x4-1x2x3)+(3x4x5-2x3x4)+.+(999x1000x1001-998x999x1000)]/3 =[999x1000x1001]/3 =333333000
迟友坚1919求和:1X2+2X3+3X4+.+nX(n+1)? -
宦通翁15720698322 ______[答案] 原式:f(x)=n(n+1)=1/3((n+2)(n+1)n-n(n-1)(n+1)) 运用数列的等差中的一些定理求出来,得1/3n(n+1)(n+2)