# Let an = 1 n(n+1): Compute a1; a1 +a2; a1 +a2 +a3; a1 +a2 +a3 +a4. Then guess a1 + a2 +…+ an

The mathematical proving technique known as mathematical induction is used to demonstrate a particular claim for any well-organized set. It is typically used to support conclusions or prove claims that are expressed in terms of n, where n is a natural number. The idea of mathematical induction aids in the proof of mathematical conclusions and theorems for all natural numbers. One method for proving certain algebraic propositions that are expressed in terms of n, a natural number, is the concept of mathematical induction.

**Method of Solving Using Mathematical Induction**

Let us assume a statement, referred to by, P(n), where n is a natural number. In order to verify the validity of P(n) for every n, the following principle can be used.

Step 1:Verify if the given statement is true for n = 1.

Step 2:Assume that the given statement P(n) is also true for n = k, for any positive integer k .

Step 3:Prove that the statement holds for P(k+1) for any positive integer k.Make a conclusion then, P(n) is true for all n natural numbers.

**Problem: Let a**_{n} = 1 n(n+1): Compute a1; a_{1} + a_{2}; a_{1} +a_{2} +a_{3}; a_{1} + a_{2} + a_{3} + a_{4}. Then guess a_{1} + a_{2 }+ …… + a_{n} for any natural number n. Use both mathematical induction and a direct approach to verify your guess.

_{n}= 1 n(n+1): Compute a1; a

_{1}+ a

_{2}; a

_{1}+a

_{2}+a

_{3}; a

_{1}+ a

_{2}+ a

_{3}+ a

_{4}. Then guess a

_{1}+ a

_{2 }+ …… + a

_{n}for any natural number n. Use both mathematical induction and a direct approach to verify your guess.

**Solution:**

The problem statement involves the calculation of a_{1}, a_{1}+a_{2}, a_{1}+a_{2}+a_{3,} and a_{1}+a_{2}+a_{3}+a_{4}.

**Using Direct Method**

Let us make an assumption that the a

_{1 }+ a_{2 }+â€¦. + a_{n}is any natural numberGiven : a

_{n}=1n (n+1)Take n =1, 2, 3, 4 to get the value of a

_{1}, a_{2}a_{3,}and a_{4}. So,Calculating the values, we obtain,

a

_{1}= 1 x 1(1+1)

= 2 ……(I)a

_{2}= 1 x 2(2+1)

= 6……..(II)a

_{3}= 1 x 3 (3+1)

= 12……(III)And,

a

_{4}= 1 x 4 (4+1)

=20 ……(IV)Solving for a

_{1}, we have,a

_{1}= 2a

_{1}+ a_{2}= 2 + 6

= 8……(V)a

_{1}+ a_{2}+ a_{3}= 2 + 6 + 12

= 20……(VI)a

_{1}+ a_{2}+ a_{3}+ a_{4}= 2 + 6 + 12 + 20

= 40……..(VII)This can also be expressed in the form of,

a

_{1}+ a_{2}+ a_{3}+ a_{4 }+ … + a_{n}= a_{1}+ a_{2}+ …. a_{n}

**Using Mathematical Induction**

Given: a

_{n}= 1 n(n+1)Let us assume,

S(n) = a

_{1}+ a_{2}+ a_{3}+ ……. + a_{n}_{ }= 1(2) + 2(3) + 3(4) + â€¦ + n(n+1)This is equivalent to , S(n) = [n(n+1)(n+2)] / 3……..[eq. A]

For n = 1,

S(1) = a

_{1}

=1(2) = 2On substituting the values in eq. A

S(1) = 1(1+1)(1+2)/3

= 2Therefore, 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

## Solved Examples

**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 + â€¦.. + (4n âˆ’ 2) = 2n ^{2} 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 Ã— 1

^{2}= 2Thus, P(1) is true.

Consider P(k) is true,

P(k) = 2 + 6 + 10 + â€¦.. + (4k âˆ’ 2) = 2k

^{2}For, P (k + 1),

LHS = 2 + 6 + 10 + â€¦.. + (4k âˆ’ 2) + (4(k + 1) âˆ’ 2)

= 2k^{2}+ (4k + 4 âˆ’ 2)

= 2k^{2}+ 4k + 2

= (k+1)^{2}^{ }= RHSHere 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 3 ^{n} âˆ’ 1 is divisible by 2 by using the principle of mathematical induction.**

**Solution:**

Here we have to prove that, 3

^{n}âˆ’ 1 is divisible by 2 for all the positive integersNow,

Assume, n= 1

So, P(1) = 3

^{1}âˆ’ 1 = 2Thus, it is divisible by 2. Therefore, P(1) is true.

Consider that P(k) is true or 3

^{k}âˆ’ 1 is divisible by 2.Thus C becomes,

LHS = 3

^{(k + 1)}âˆ’ 1= 3

^{k}Ã— 3 âˆ’ 1= 3

^{k}Ã— 3 âˆ’ 3 + 2= 3(3

^{k}âˆ’ 1) + 2Now, (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 3

^{n}– 1 is divisible by 2

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