Let an = 1 n(n+1): Compute a1; a1 +a2; a1 +a2 +a3; a1 +a2 +a3 +a4. Then guess a1 + a2 +…+ 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 a_n = 1 n(n+1): Compute a1; a₁ + a₂; a₁ +a₂ +a₃; a₁ + a₂ + a₃ + a₄. Then guess a₁ + a₂ + …… + a_n for any natural number n. Use both mathematical induction and a direct approach to verify your guess.×

Solution:

roblem statement involves the calculation of a₁, a₁+a₂, a₁+a₂+a_3, and a₁+a₂+a₃+a₄. 

Using Direct Method

Let us make an assumption that the a₁ + a₂ +…. + a_n is any natural number

Given:

• a_n =1n (n+1)

Take n =1, 2, 3, 4 to get the value of a₁, a₂ a_3, and a₄. So,

Calculating the values, we obtain

• a₁ = 1 × 1(1+1) = 2 ……(i)
• a₂ = 1 × 2(2+1) = 6……..(ii)
• a₃ = 1 × 3 (3+1) = 12……(iii)
• a₄ = 1 × 4 (4+1) =20 ……(iv)

Solving for a₁, we have, 

a₁ = 2 

a₁ + a₂ = 2 + 6 = 8……(v)

a₁ + a₂ + a₃ = 2 + 6 + 12 = 20……(vi)

a₁ + a₂ + a₃ + a₄ = 2 + 6 + 12 + 20 = 40……..(vii)

This can also be expressed in the form of, 

a₁ + a₂ + a₃ + a₄ + … + a_n= a₁ + a₂ + …. a_n

Using Mathematical Induction

Given:

• a_n = 1 n(n+1)

Let us assume, 

S(n) = a₁ + a₂ + a₃ + ……. + a_n = 1(2) + 2(3) + 3(4) + … + n(n+1)

S(n) = [n(n+1)(n+2)] / 3……..[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 a_n = 1 n(n+1) holds true by the Principle of Mathematical Induction.

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Example 1: Find the sum of the first 20 terms for the n^th term, a_n = n(n+1).

Solution 

As calculated above, for the given n^th 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, a_n = 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,

⇒ S = 40

⇒ n (n+1) (n+2)/3 = 40

⇒ n (n+1) (n+2) = 120

Solving the above equation we get,

⇒ 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 ……….. (I)

Let us now prove, 

P(k +1) is true whenever P(k) is true.

On multiplication of I by 2, we obtain, 

⇒ 2. 2k > 2k

On further solving, we get,\

⇒ 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 + ….. + (4n − 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 × 1² = 2

Thus, P(1) is true.

Consider P(k) is true,

P(k) = 2 + 6 + 10 + ….. + (4k − 2) = 2k²

For, P (k + 1),

LHS = 2 + 6 + 10 + ….. + (4k − 2) + (4(k + 1) − 2)

= 2k² + (4k + 4 − 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 − 1 is divisible by 2 by using the principle of mathematical induction.

Solution:

Here we have to prove that, 3ⁿ − 1 is divisible by 2 for all the positive integers

Assume, n= 1

So, P(1) = 3¹ − 1 = 2 

Thus, it is divisible by 2. Therefore, P(1) is true.

Consider that P(k) is true or 3^k − 1 is divisible by 2.

Thus C becomes,

LHS = 3^{(k + 1)} − 1 

= 3^k × 3 − 1 

= 3^k × 3 − 3 + 2 

= 3(3^k − 1) + 2

Now, (3^k – 1) and 2 both of them are divisible by 2, hence,  P(k+1) is true.

So, by the Principle of Mathematical Induction 3ⁿ – 1 is divisible by 2