急...數學歸納法証明???

Let n=1 then
LHS:
1+2+…+2n=1+2=3
RHS:
n(2n+1)=1*(2*1+1)=3

therefore for n=1 LHS=RHS, it is true
let n=2 then
LHS:
1+2+…+2n=1+2+3+4=10
RHS:
n(2n+1)=2*(2*2+1)=10
therefore for n=2 LHS=RHS, it is true

now assume it is true for n=a where a is any positive integer, i.e. 1+2+…+2a=a(2a+1)
when n=a+1 then
LHS
1+2+…+2a+2a+1+2(a+1)=a(2a+1)+2a+1+2(a+1)
=(a+1)(2a+1)+2(a+1)
=(a+1)(2a+1+2)
=(a+1)(2(a+1)+1)
RHS:
(n)(2n+1)=(a+1)(2(a+1)+1)
LHS=RHS

Therefore it is also true for n=a+1

Let P(n) be the statemnet 1+2+....+2n =n(2n+1)

for n=1
L.H.S=1+2=3

R.H.S= 1(2+1)
=3

so P(1) is true

Assume P(k) is true for an integer k
for n=k+1
P(k+1):
L.H.S =1+2+......2(k)+(2k+1)+2(k+1)
=k(2k+1)+(2k+1)+2(k+1)
=(2k+1)(k+1)+2(k+1)
=(k+1)(2k+1+2)
=(k+1)[2(k+1)+1]
=R.H.S
so P(k+1) is also true.

By M.I. P(n) is true for all integer n.