If n is odd,then (1+3+5+7+..........to n terms) is equal to
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Answered by
8
let No. of terms in (1+3+5.....n) be x
then,
last term of Ap will be-
n = 1 + (x-1)2
X =
Now u can use general Formula to find the Sum
Sn =
where n = no. of terms in Ap
Replacing n with (n+1)/2
a with 1 , d with 2
.
then,
last term of Ap will be-
n = 1 + (x-1)2
X =
Now u can use general Formula to find the Sum
Sn =
where n = no. of terms in Ap
Replacing n with (n+1)/2
a with 1 , d with 2
.
Answered by
1
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
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