Question

If n is odd,then (1+3+5+7+..........to n terms) is equal to

Answer 1

let No. of terms in (1+3+5.....n) be x
then,
last term of Ap will be-
n = 1 + (x-1)2
X =
$\frac{n - 1}{2} + 1 = \frac{n + 1}{2}$
Now u can use general Formula to find the Sum

Sn =
$\frac{n(2a + (n - 1)d)}{2}$
where n = no. of terms in Ap

Replacing n with (n+1)/2
a with 1 , d with 2
.
$(n + 1)(2 + ( \frac{n + 1}{2} - 1)2$

Answer 2

Answer:

let No. of terms in (1+3+5.....n) be x

then,

last term of Ap will be-

n = 1 + (x-1)2

X =

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

2n−1+1= 2n+1

Now u can use general Formula to find the Sum

Sn =

\frac{n(2a + (n - 1)d)}{2}

2n(2a+(n−1)d)

where n = no. of terms in Ap

Replacing n with (n+1)/2

a with 1 , d with 2

.

(n + 1)(2 + ( \frac{n + 1}{2} - 1)2(n+1)(2+( 2n+1−1)2