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Question: If the \(n ^ { t h }\) term of an A.P. be\(( 2 n - 1 )\), then the sum of its first \(n\) terms wi...

If the nthn ^ { t h } term of an A.P. be(2n−1)( 2 n - 1 ), then the sum of its first nn terms will be.

A

n2−1n ^ { 2 } - 1

B

(2n−1)2( 2 n - 1 ) ^ { 2 }

C

n2n ^ { 2 }

D

n2+1n ^ { 2 } + 1

Answer

n2n ^ { 2 }

Explanation

Solution

Given that Tn=2n−1T _ { n } = 2 n - 1

First term a=2.1−1=1a = 2.1 - 1 = 1

Second term b=2.2−1=3b = 2.2 - 1 = 3

Third term c=2.3−1=5c = 2.3 - 1 = 5

Therefore sequence is .

Now sum of the first nn terms is Sn=n2[a+l]S _ { n } = \frac { n } { 2 } [ a + l ]

=n2[1+2n−1]=n2⋅2n=n2= \frac { n } { 2 } [ 1 + 2 n - 1 ] = \frac { n } { 2 } \cdot 2 n = n ^ { 2 }

Aliter : Since Tn=2n−1T _ { n } = 2 n - 1

⇒Sn=ΣTn=2Σn−Σ1=n(n+1)−n=n2\Rightarrow S _ { n } = \Sigma T _ { n } = 2 \Sigma n - \Sigma 1 = n ( n + 1 ) - n = n ^ { 2 }