作业帮1/(1*4)+1/(4*7)+1/(7*10)+……+1/[（3n-2)(3n+1)]求和
来源：学生作业帮助网 编辑：作业帮 时间：2024/08/03 05:17:15
![1/(1*4)+1/(4*7)+1/(7*10)+……+1/[（3n-2)(3n+1)]求和](/uploads/image/z/489197-29-7.jpg?t=1%2F%281%2A4%29%2B1%2F%284%2A7%29%2B1%2F%287%2A10%29%2B%E2%80%A6%E2%80%A6%2B1%2F%5B%EF%BC%883n-2%29%283n%2B1%29%5D%E6%B1%82%E5%92%8C)
x)302&Z`\@SQ2 0506Ԍ} "}5n6[c
C]C} 6aJB՚PZ`(by ~qAb / Zb
1/(1*4)+1/(4*7)+1/(7*10)+……+1/[（3n-2)(3n+1)]求和
1/(1*4)+1/(4*7)+1/(7*10)+……+1/[（3n-2)(3n+1)]求和
1/(1*4)+1/(4*7)+1/(7*10)+……+1/[（3n-2)(3n+1)]求和
1/(1*4)+1/(4*7)+1/(7*10)+……+1/[（3n-2)(3n+1)]
=3(1-1/4+1/4-1/7+...+1/（3n-2)-1/(3n+1))
=3*(1-1/(3n+1))
=9n/3n+1
1、1/2、1/4、1/7
1/4+1/7-0.25
25.6-2/7-4/7-1/7
计算(7+1)(7^2+1)(7^4+1)(7^8+1)(7^16+1)
6(7+1)(7^2+1)(7^4+1)(7^8+1)(7^16+1)
-8×7/2×(-4/7)×1/4
1/4+1/4*1/7+1/7*1/10+...1/25*1/28=?
计算6(7+1)(7^+1)(7^4+1)(7^8+1)+1,
计算：6(7+1)(7^2+1)(7^4+1)(7^8+1)+1
1/7*2+1/7*4+1/7*8+1/7*16+1/7*32+1/7*64=?用拆项法解
1/1*4+1/4*7+1/7*10+.+1/2005*2008 1/1*4+1/4*7+1/7*10+.+1/2005*2008
1/1*4+1/4*7+1/7*10+.+1/52*55*+1/55*58=?
1/1×4+1/4×7+1/7×10+.+1/52×55+1/55×58
1/1*4+1/4*7+1/7*10+1/10*13+~+1/97*100
1如何计算1*1/4+4*1/7+7*1/10+...+2002*1/2005
计算1/1*4+1/4*7+1/7*10+...+1/2005*2008+1/2008*2011
1/1*4+1/4*7+1/7*10+1/10*13+…+1/298*301
1/1×4+1/4×7+1/7×10+...+1/22×25+1/25×28