n(n+1)=⅓×[n(n+1)(n+2)-(n-1)n(n+1)]
1×2+2×3+...+n(n+1)
=⅓×[1×2×3-0×1×2+2×3×4-1×2×3+...+n(n+1)(n+2)-(n-1)n(n+1)]
=⅓n(n+1)(n+2)
解法二：
n(n+1)=n²+n
1×2+2×3+...+n(n+1)
=(1²+2²+...+n²)+(1+2+...+n)
=n(n+1)(2n+1)/6 +n(n+1)/2
=[n(n+1)/6](2n+1+3)
=n(n+1)(2n+4)/6
=⅓n(n+1)(n+2)本回答被网友采纳
1X2十2X3+……n(n+1)=?
n(n+1)=n²+n 1×2+2×3+...+n(n+1)=(1²+2²+...+n²)+(1+2+...+n)=n(n+1)(2n+1)\/6 +n(n+1)\/2 =[n(n+1)\/6](2n+1+3)=n(n+1)(2n+4)\/6 =⅓n(n+1)(n+2)
求解,Sn=1x2+2x3+3x4+…+n(n+1)=?跪求,要过程
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！！
求和:1X2+2X3+3X4+.+nX(n+1)?
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)
求和:1X2+2X3+3X4+...+nX(n+1)?
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)满意请采纳
1X2+2X3+3X4+...NX(N+1)规律
an = n(n+1)即 an = n^2 + n n(n+1)(2n+1)其中1^2 + 2^2 + 3^2 + ... + n^2 = --- 6 n(n+1)1+2+3+...+n = --- 2 所以 1X2+2X3+3X4+...NX(N+1)N(N+1)(2N+1) N(N+1)= --- + --- 6 2 N(N+1)(2N+4)= --- 6 N...
1x2+2x3+3x4+4x5+...+n(n+1)等于多少?急救!!
在这里我只给你点提示：公式一 n(n+1)=n^2+n 公式二 1^2+2^2+3^3+…+n^2=n(n+1)(2n+1)\/6 公式三 1+2+3+…+n=n(n+1)\/2 上面的兄弟已经解答出来了。很正确，不过怕您不会用公式二。关于公式二的证明：1^2+2^2+3^2+……+n^2=n(n+1)(2n+1)\/6 利用立方差公式 ...
1x2 2x3 3x4 ... n(n 1)=?过程
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
1x2+2x3+3x4+4x5+···n(n+1)=___(n为自然数)
n(n+1)=n^2+nS(n)=(1+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 =n(n+1)(n+2)\/3
简算1x2+2x3+3x4+4x5+...+n(n+1)
1x2+2x3+3x4+4x5+...+n(n+1)=(1^2+2^2+……n^2)+(1+2+3+……n)=n(n+1)(2n+1)\/6+(1+n)xn\/2 =n(n+1)(n+2)\/3
1x2十2x3+3x4……十nx(n十1)
1x2十2x3+3x4……十nx(n十1)=(1^2+2^2+...+n^2)+(1+2+...+n)=n(n+1)(2n+1)\/6+n(n+1)\/2 =n(n+1)(n+2)\/3 如果不懂，请追问，祝学习愉快！
