Question

if n is odd ,then(1+3+5+7+....to n terms)is equal to.
Full answer with explanation

Answer 1

We have:

1 + 3 + 5 + 7 + ......... + to n terms

where, n is odd number

We have to find, 1 + 3 + 5 + 7 + ......... + to n terms = ?

Solution:

1 + 3 + 5 + 7 + ......... + to n terms

The given series are in AP.

Here, first term (a) = 1, common difference (d) = 3 - 1 = 2

and number of terms (n) = n

We know that,

The sum of up to n terms

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

⇒ $S_{n} =\dfrac{n}{2} [2(1)+(n-1)2]$

⇒ $S_{n} =\dfrac{n}{2}$ (2 + (n - 1)2)

⇒ $S_{n} =\dfrac{n}{2}$ (2 + 2n - 2)

⇒ $S_{n} =\dfrac{n}{2}$ (2n)

⇒ $S_{n} =n^2$

∴ 1 + 3 + 5 + 7 + ......... + to n terms = $n^2$

Thus, 1 + 3 + 5 + 7 + ......... + to n terms = $n^2$

Answer 2

Given :

n is an odd number .

To Find :

Sum of the series (1+3+5+7+.... to n terms )  = ?

Solution :

∴ It is given that n is an odd number , so the total no of termsin the given series is :

$= \frac{n+1}{2}$

Now , we can see that the given series is an arithmetic progression , where :

First term , a = 1

And common difference , d = 2

∴We know that the formula for the sum of n terms of an AP is given as :

$S_{n} = \frac{n}{2}\times [ 2a + ( n-1) \ties d ]$

So on using the above formula to find the sum of the given series  we get :

$S_{n} =(\frac{n+1}{2})\times [2\times 1 + (\frac{n+1}{2} -1) \times 2]$

$S_{n} =( \frac{n+ 1}{2})\times [ 2 + (\frac{n+1-2}{2} )\times 2]$

$S_{n} =( \frac{n+1}{2})\times [ 2 + ( n-1)]\\ \\S_{n} = (\frac{n+1}{2} )\times ( n+1)$

$S_{n}= \frac{(n+1)^{2} }{2}$

So the sum of given series upto n terms is $S_{n}= \frac{(n+1)^{2} }{2}$ .