Question

数学题，求详细过程。

求和。
(1)(a-1)+(a²-2)+…+(a∧n-n)(a≠0)

(2)1+3×3+5×3²+…+(2n-1)×3∧n-1

(3)1/1×4+1/4×7+1/7×10+…+1/(3n-2)(3n+1)

(4)1/1+√2+1/√2+√3+…+1/√n+√n+1

Answer 1

(1)(a-1)+(a²-2)+…+(a∧n-n)(a≠0)
=a+a²+…+a∧n-（1+2+……n）
=(a∧n-a)/(a-1)-n(n+1)/2

(2)1+3×3+5×3²+…+(2n-1)×3∧n-1
Sn=1+3*3+5*3^2+...+(2n-1)*3^(n-1),得

3Sn=1*3+3*3^2+...+(2n-1)*3^n
错位相减得
-2Sn=1+2*3+2*3^2+2*3^3+...+2*3^(n-1)-(2n-1)*3^n
=2(3^n-n*3^n-1)
Sn=(n-1)*3^n+1

1+3×3+5×3²+…+(2n-1)×3∧n-1=(n-1)*3^n+1

(3)1/1×4+1/4×7+1/7×10+…+1/(3n-2)(3n+1)
=[1-1/4+1/4-1/7+1/7-1/10+……+1/(3n-2)-1/(3n+1)]/3
=[1-1/(3n+1)]/3
=n/(3n+1)

(4)1/(1+√2)+1/(√2+√3)+…+1/(√n+√n+1)
=√2-1+√3-√2+…+√n+1-√n
=√(n+1)-1

Answer 2

1等比数列之和+ 等差数列之和
2错位相减 等比数列之和
3裂项相消
4分母有理化