Answer
a) 2+4+6+.... +2n=n (n+1)
b) BASIC STEP:
it is given that the n is positive integer then,
put n=1
2=n (n+1) = 1 (1+1) =2
(The left-hand side of this equation is 2 because 2 is the sum of the first a positive integer. The right-hand side is found by substituting 2 for n in n (n+1)).
INDUCTIVE STEP:
For the inductive hypothesis we assume that P(k)
2+4+6+…. +2k=k (k+1)
Under the assumption, it must be show that P(k+1) is true, namely, that.
2+4+6… +2(k+1) =(k+1) (k+2)
When we add 2(k+1) to both side of the equation in P(K), we obtain.
2+4+6+… +2k+2(k+1) = k (k+1) +2(k+1)
[By the inductive hypothesis]
= (k+1) (k+2)
We have completed the basis step and the inductive step, so by mathematical inductive we know that P(n) is true for all positive integers n.
That is, we have proven that.
2+4+6+.... +2n=n (n+1)
Work Step by Step
a) 2+4+6+.... +2n=n (n+1)
b) BASIC STEP:
it is given that the n is positive integer then,
put n=1
2=n (n+1) = 1 (1+1) =2
(The left-hand side of this equation is 2 because 2 is the sum of the first a positive integer. The right-hand side is found by substituting 2 for n in n (n+1) ).
INDUCTIVE STEP:
For the inductive hypothesis we assume that P(k)
2+4+6+…. +2k=k (k+1)
Under the assumption, it must be show that P(k+1) is true, namely, that
2+4+6… +2(k+1)=(k+1) (k+2)
When we add 2(k+1) to both side of the equation in P(K), we obtain
2+4+6+… +2k+2(k+1) = k (k+1)+2(k+1)
[By the inductive hypothesis]
= (k+1)(k+2)
We have completed the basis step and the inductive step, so by mathematical inductive we know that P(n) is true for all positive integers n.
That is, we have proven that.
2+4+6+.... +2n=n (n+1)
