Answer
a) 1/2, 2/3 and n/n+1
b) BASIC STEP:
it is given that the n is positive integer then,
put n=1
1/2 =1/1+1 =1/2
(The left-hand side of this equation is 1/2 because 1/2 is the sum of the first a positive integer. The right-hand side is found by substituting 1/2 for n in n/n+1
INDUCTIVE STEP:
For the inductive hypothesis we assume that $P(k $)
$1/1.2+1/2.3+... +1/k (k+1) = k/k+1$
Under the assumption, it must be show that$ P(k+1)$ is true, namely, that.
$1/1.2+1/2.3+... +1/(k+1) (k+2) = k+1/k+2$
When we add 1/(k+1) (k+2) to both side of the equation in P(K), we obtain.
$1/1.2+1/2.3+... +1/k (k+1) + 1/(k+1) (k+2) = k/k+1 + 1/(k+1) (k+2) $
[By the inductive hypothesis]
= $k(k+2) +1/(k+1) (k+2)$
=$k2+2k+1/(k+1) (k+2)$
=$k+1/k+2$
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.
$1/1.2+1/2.3+... +1/n (n+1) $=$ n/n+1$
Work Step by Step
a) 1/2, 2/3 and n/n+1
b) BASIC STEP:
it is given that the n is positive integer then,
put n=1
1/2 =1/1+1 =1/2
(The left-hand side of this equation is 1/2 because 1/2 is the sum of the first a positive integer. The right-hand side is found by substituting 1/2 for n in n/n+1
INDUCTIVE STEP:
For the inductive hypothesis we assume that P(k)
1/1.2+1/2.3+... +1/k (k+1) = k/k+1
Under the assumption, it must be show that P(k+1) is true, namely, that.
1/1.2+1/2.3+... +1/(k+1) (k+2) = k+1/k+2
When we add 1/(k+1) (k+2) to both side of the equation in P(K), we obtain.
1/1.2+1/2.3+... +1/k (k+1) + 1/(k+1) (k+2) = k/k+1 + 1/(k+1) (k+2)
[By the inductive hypothesis]
= k(k+2) +1/(k+1) (k+2)
=k2+2k+1/(k+1) (k+2)
=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.
1/1.2+1/2.3+... +1/n (n+1) = n/n+1
