Prove Summation Identities using Mathematical Induction examples with solutions
Question 1:
Prove the following statement by the principle of mathematical induction:
, n ∈ N.
Solution:
We can write,
Step 1:
We first show that the basis for induction P(1) is true, that is, P(n) is true for .
Left Side = 1,
Right Side .
Both sides of the statement are equal for . Hence, P(1) is true, that is, P(n) is true for .
Step 2:
Assume that P(k) is true for some natural number k, that is,
We need to prove that P(k +1) is also true.
Because P(k) is true, so we have,
Add to both sides:
.
That is, we get
.
Thus, P(k +1) is true whenever P(k) is true.
Therefore, from the principle of mathematical induction, the statement:
is true for all natural number n.
Question 2:
Prove the following statement by the principle of mathematical induction:
, n ∈ N.
Solution:
We can write,
Step 1:
We first show that the basis for induction P(1) is true, that is, P(n) is true for .
Left Side ,
Right Side .
Both sides of the statement are equal for . Hence, P(1) is true, that is, P(n) is true for .
Step 2:
Assume that P(k) is true for some natural number k, that is,
We need to prove that P(k +1) is also true.
Because P(k) is true, so we have,
Add to both sides:
.
That is, we get
.
Thus, P(k +1) is true whenever P(k) is true.
Therefore, from the principle of mathematical induction, the statement:
is true for all natural number n.
Tags: mathematical induction examples with solutions, prove summation identities by induction
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