Simillar Examples
Example 1: Find the sum of the first 20 terms for the nth term, an = n(n+1).
Solution
As calculated above, for the given nth term the sum is given by,
S(n) = n (n+1) (n+2)/3
Putting n = 20 we get,
S(20) = (20)(21)(23)/3
= 3220
Example 2. Find the value of n for the nth term, an = n(n+1) if the sum of the first n terms is 40.
Solution:
As calculated above, for the given nth term the sum is given by,
S(n) = n (n+1) (n+2)/3
We are given that,
β S = 40
β n (n+1) (n+2)/3 = 40
β n (n+1) (n+2) = 120
Solving the above equation we get,
β n = 4
Example 3. Prove that 2n > n for all positive integers n.
Solution:
Let us assume, that P(n): 2n > n
For n = 1,
We obtain 2n = 2. Since, 2n > n, that is 2 > 1
Therefore, P(1) holds true.
Let us assume now that P(k) is true for any positive integer k, for instance,
2k > k β¦β¦β¦.. (I)
Let us now prove,
P(k +1) is true whenever P(k) is true.
On multiplication of I by 2, we obtain,
β 2. 2k > 2k
On further solving, we get,\
β 2 k + 1 > 2k = k + k > k + 1
Hence, P(k + 1) is true when P(k) is true.
Therefore, this statement holds for all natural numbers n.
Example 4. For all the positive integers prove that the equation 2 + 6 + 10 + β¦.. + (4n β 2) = 2n2 is true by using the principle of mathematical induction.
Solution:
By using the statement formula
For, n = 1 or P(1),
Here, LHS = 2 and RHS = 2 Γ 12 = 2
Thus, P(1) is true.
Consider P(k) is true,
P(k) = 2 + 6 + 10 + β¦.. + (4k β 2) = 2k2
For, P (k + 1),
LHS = 2 + 6 + 10 + β¦.. + (4k β 2) + (4(k + 1) β 2)
= 2k2 + (4k + 4 β 2)
= 2k2 + 4k + 2
= (k+1)2
= RHS
Here we have proved that P(k+1) is also true.
Therefore, by the principle of mathematical induction the given statement is true for all positive integers.
Hence proved.
Example 5. Find that for all the positive integers, the expression 3n β 1 is divisible by 2 by using the principle of mathematical induction.
Solution:
Here we have to prove that, 3n β 1 is divisible by 2 for all the positive integers
Assume, n= 1
So, P(1) = 31 β 1 = 2
Thus, it is divisible by 2. Therefore, P(1) is true.
Consider that P(k) is true or 3k β 1 is divisible by 2.
Thus C becomes,
LHS = 3(k + 1) β 1
= 3k Γ 3 β 1
= 3k Γ 3 β 3 + 2
= 3(3k β 1) + 2
Now, (3k β 1) and 2 both of them are divisible by 2, hence, P(k+1) is true.
So, by the Principle of Mathematical Induction 3n β 1 is divisible by 2
Let an = 1 n(n+1): Compute a1; a1 +a2; a1 +a2 +a3; a1 +a2 +a3 +a4. Then guess a1 + a2 +β¦+ an
The solution for the given statement is achieved using, directly solving the expression or by mathematical induction. The detailed solution for the same is added below:
Problem: Let an = 1 n(n+1): Compute a1; a1 + a2; a1 +a2 +a3; a1 + a2 + a3 + a4. Then guess a1 + a2 + β¦β¦ + an for any natural number n. Use both mathematical induction and a direct approach to verify your guess.Γ
Solution:
roblem statement involves the calculation of a1, a1+a2, a1+a2+a3, and a1+a2+a3+a4.
Using Direct Method
Let us make an assumption that the a1 + a2 +β¦. + an is any natural number
Given:
- an =1n (n+1)
Take n =1, 2, 3, 4 to get the value of a1, a2 a3, and a4. So,
Calculating the values, we obtain
- a1 = 1 Γ 1(1+1) = 2 β¦β¦(i)
- a2 = 1 Γ 2(2+1) = 6β¦β¦..(ii)
- a3 = 1 Γ 3 (3+1) = 12β¦β¦(iii)
- a4 = 1 Γ 4 (4+1) =20 β¦β¦(iv)
Solving for a1, we have,
a1 = 2
a1 + a2 = 2 + 6 = 8β¦β¦(v)
a1 + a2 + a3 = 2 + 6 + 12 = 20β¦β¦(vi)
a1 + a2 + a3 + a4 = 2 + 6 + 12 + 20 = 40β¦β¦..(vii)
This can also be expressed in the form of,
a1 + a2 + a3 + a4 + β¦ + an= a1 + a2 + β¦. an
Using Mathematical Induction
Given:
- an = 1 n(n+1)
Let us assume,
S(n) = a1 + a2 + a3 + β¦β¦. + an = 1(2) + 2(3) + 3(4) + β¦ + n(n+1)
S(n) = [n(n+1)(n+2)] / 3β¦β¦..[eq. A]
For n = 1,
S(1) = a1 =1(2) = 2
On substituting the values in eq. A
S(1) = 1(1+1)(1+2)/3 = 2
Therefore, the statement holds for n = 1
Let us assume the statement holds for k
Now, let us prove that the statement holds for k + 1,
Suppose, S(k) = [k(k+1)(k+2)] / 3
Then,
S(k+1) = S(k) + (k+1)(k+2)
= k(k+1)(k+2)/3 + (k+1)(k+2)
= (k+1)(k+2)(k/3 + 1)
S(k+1) = (k+1)(k+2)(k+3)/3
Thus, S(k + 1) is true.
So, the statement an = 1 n(n+1) holds true by the Principle of Mathematical Induction.