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

(a) $(n^{2} + 1)$

(b) $(n^{2} - 1)$

(c) $n^{2}$

(d) $(2 n^{2} + 1)$

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Solution

Consider, $1 + 3 + 5 + 7 + . . . . . +$ to $n$ terms
Here, $a = 1, d = 5 - 3 = 3 - 1 = 2$ , therefore, the given series is an AP.
Therefore, using sum of $n$ terms of an AP, we get,
$\Rightarrow S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$\Rightarrow S_{n} = \frac{n}{2} [2 (1) + (n - 1) (2)]$
$\Rightarrow S_{n} = n [1 + (n - 1)]$
$\Rightarrow S_{n} = n^{2}$
Therefore, sum of $n$ terms of an odd series is $n^{2}$ .

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If n is odd, then 1+3+5+7+.....+ to n terms is equal to (a) (n2+1) (b) (n2−1) (c) n2 (d) (2n2+1)

Consider, 1+3+5+7+.....+ to n termsHere, a=1,d=5−3=3−1=2, therefore, the given series is an AP.Therefore, using sum of n terms of an AP, we get,⇒Sn=n2[2a+(n−1)d]⇒Sn=n2[2(1)+(n−1)(2)]⇒Sn=n[1+(n−1)]⇒Sn=n2Therefore, sum of n terms of an odd series is n2.

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

(a) $(n^{2} + 1)$

(b) $(n^{2} - 1)$

(c) $n^{2}$

(d) $(2 n^{2} + 1)$