f4 amath-MI

As follows:

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Let P(n) be the proposition of "1*2+2*3+2^2*4+...+2^N-1*(N+1)=2^N*(N) "for all positive integer n.
for n=1,
LHS=2^(1-1)*(1+1)=1*2=2=2^1*1=RHS
therefore P(1) is true.

Assume P(k) is true,
i.e, 1*2+2*3+2^2*4+...+2^(k-1)*(k+1)=2^k*(k)

for P(k+1),
LHS=1*2+2*3+2^2*4+...+2^(k-1)*(k+1)+2^(k+1-1)*(k+1+1)
=2^k*(k)+2^(k+1-1)*(k+1+1)
=2^k*(k)+2^(k)*(k+2)
=2^k*(k+k+2)
=2^k*2(k+1)
=2^(k+1)*(k+1)
=RHS
therefore, P(k+1) is true.
By MI, P(n) is true for all positive integer.

Let the preposition P(n) be
"1*2+2*3+2^2*4+...+2^N-1*(N+1)=2^N*(N) FOR ALL POSITIVE INTEGERS N"

For n=1,
L.H.S.= 1*2=2
R.H.S.= 2^1(1)=2=L.H.S.
Therefore, P(1) is true.

Assume P(n) is true for any positive integers n.
that is
1*2+2*3+2^2*4+...+2^N-1*(N+1)=2^N*(N) for any positive integers n

Consider P(n+1),
L.H.S.= 1*2+2*3+2^2*4+...+2^N-1*(N+1) + 2^(N)*(N+2)
= 2^N*(N) + 2^(N)*(N+2)
=2^N*(N+N+2)
=2^N*(2N+2)
=2^N(2)*(N+1)
=2^(N+1)*(N+1) = R.H.S.

Therefore,P(n+1) is true.
By MI, the preposition P(n) is true for all positive integers n.