Question

if n is odd then (1+3+5+7+____ to n terms) is equal too​

Answer 1

Answer:

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

=

2

n

[2a+(n−1)d]

⇒ S_{n} =\dfrac{n}{2} [2(1)+(n-1)2]S

n

=

2

n

[2(1)+(n−1)2]

⇒ S_{n} =\dfrac{n}{2}S

n

=

2

n

(2 + (n - 1)2)

⇒ S_{n} =\dfrac{n}{2}S

n

=

2

n

(2 + 2n - 2)

⇒ S_{n} =\dfrac{n}{2}S

n

=

2

n

(2n)

⇒ S_{n} =n^2S

n

=n

2

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

2

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

2