1.3 + 3.5 + 5.7 + … + (2n – 1) (2n + 1) = $$\dfrac{n(4n^2 + 6n – 1)}{3}$$

Asked by Aaryan | 1 year ago |  92

##### Solution :-

Let us check for n = 1,

P (1): (2.1 – 1) (2.1 + 1) = $$\dfrac{ 1(4.1^2 + 6.1 -1)}{3}$$

1×3 = $$\dfrac{ 1(4+6-1)}{3}$$

3 = 3

P (n) is true for n = 1.

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): 1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) =$$\dfrac{ k(4k^2 + 6k – 1)}{3}$$ … (i)

So,

1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + (2k + 1) (2k + 3)

Now, substituting the value of P (k) we get,

= $$\dfrac{ k(4k^2 + 6k – 1)}{3}$$ + (2k + 1) (2k + 3) by using equation (i)

=$$\dfrac{(k+1) (4k^2 + 8k +4 + 6k + 6 – 1)] }{ 3}$$

=$$\dfrac{(k+1) 4(k+1)2 + 6(k+1) -1 }{3}$$

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

Answered by Aaryan | 1 year ago

### Related Questions

#### Given an example of a statement P (n) such that it is true for all n ϵ N.

Given an example of a statement P (n) such that it is true for all n ϵ N.

#### a + (a + d) + (a + 2d) + … + (a + (n-1)d) = n/2 [2a + (n-1)d]

a + (a + d) + (a + 2d) + … + (a + (n-1)d) = $$\dfrac{n}{2}$$ [2a + (n-1)d]

#### 72n + 23n – 3. 3n – 1 is divisible by 25 for all n ϵ N

72n + 23n – 3. 3n – 1 is divisible by 25 for all n ϵ N