Prove that 1-1/2+1/3-1/4+.......+(1/2n-1)-(1/2n)=(1/n+1)+(1/n+2)+....+(1/2n),for

Question

Prove that 1-1/2+1/3-1/4+.......+(1/2n-1)-(1/2n)=(1/n+1)+(1/n+2)+....+(1/2n),for all n in N

Proof: Let P(n) be the statement 1-1/2+1/3-1/4+...+(1/2n-1)-(1/2n)=(1/n+1)+(1/n+2)+...+(1/2n),for all n in N.

Induction Basis:


Induction Step:


Thus p(n) is true for all n.
Comment: The parenthesis are only inserted for clarity as to the terms and the signs preceeding and following the terms. Prior to solving please remove all parnthesis so that you can see the problem as normal for clarity as to what I am asking you; thank you.

Explanation / Answer

proof:Let P(n) be the statement 1-1/2+1/3-1/4+...+(1/2n-1)-(1/2n)=(1/n+1)+(1/n+2)+...+(1/2n),for all n in N.

Induction basis:
step 1: check for n=1 and n=0
p(1)
LHS=1-(1/2)=1/2
RHS=1/(1+1)=1/2
P(n) is true for n=1

P(0)
LHS=1=RHS
P(n) is true for n=0

Induction Step:

step 2: assume that P(n) is true for n=k where K belongs to N
p(k)
1-1/2+1/3-1/4+......+(1/2k-1)-(1/2k)=(1/k+1)+(1/k+2)+..........+(1/2k)

step3: prove for p(n) for n=k+1 is also true
p(k+1)
RHS=(1/k+2)+(1/k+3)+..........+(1/2(k+1))
LHS=1-1/2+1/3-1/4+......+(1/2k)-(1/2(k+1))
comparing the first half of the equation with step 2 equation we get
= (1/k+1)+(1/k+2)+..........+(1/2k) +(1/2(k+1))
= RHS
p(n) is true for n=k+1

From step 1,2,3
By the principle of mathematical induction  p(n) is true for all n belongs to N.

Thus p(n) is true for all n.

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