1/(2*5)+1/(5*8)+.+1/(3n-1)(3n+2)=n/6n+4how to prove by mathematic induction?

1/(2*5)+1/(5*8)+.+1/(3n-1)(3n+2)=n/6n+4how to prove by mathematic induction?
1/(2*5)+1/(5*8)+.+1/(3n-1)(3n+2)=n/6n+4
how to prove by mathematic induction?

1/(2*5)+1/(5*8)+.+1/(3n-1)(3n+2)=n/6n+4how to prove by mathematic induction?
1/（2*5）+1/（5*8）+1/（8*11）+.+1/（(3n-1)(3n+2)
=1/3*(1/2-1/5)+1/3*(1/5-1/8)+.+1/3(1/(3n-1)-1/(3n+2))
=1/3(1/2-1/5+1/5-1/8+1/8-.+1/(3n-1)-1/(3n+2))
=1/3*(0.5-1/(3n+2)
=n/(6n+4)
综上对于任意正整数n上述等式成立.

因为
1/（2*5）+1/（5*8）+1/（8*11）+。。。。。+1/（(3n-1)(3n+2)
=1/3*(1/2-1/5)+1/3*(1/5-1/8)+........+1/3(1/(3n-1)-1/(3n+2))
=1/3(1/2-1/5+1/5-1/8+1/8-......+1/(3n-1)-1/(3n+2))
=1/3*(0.5-1/(3n+2)
=n/(6n+4)

裂项分解法

prove:
1)当n=1,1/2*5=1/(6+4).即n=1,命题成立
2)假设n=k,等式成立。
则n=k+1时，等式左边=1/(2*5)+1/(5*8)+...............+1/(3k-1)(3k+2)+1/(3k+2)(3k+5)=k/6k+4+1/(3k+2)(3k+5)=(k+1)/(6(k+1)+4)=等式右边
即n=k+1时等式也成立。
综上对于任意正整数n上述等式成立。

1/(3n-1)(3n+2) = {1/(3n-1) - 1/(3n+2)} / 3
1/(2*5)+1/(5*8)+...............+1/(3n-1)(3n+2)
=1/3[ 1/2 - 1/5 + 1/5 - 1/8 + ......+ 1/(3n-1) - 1/(3n+2) ]
=1/3[ 1/2 - 1/(3n+2) ]
= n/6n+4