Correct option is (C) 4n2 – 1
1, 3, 5, ........ forms an A.P. with common difference 2.
\(\therefore\) \(n^{th}\) term \(=a_n=a+(n-1)d\)
\(=1+(n-1)2\)
\(=2n-2+1\)
\(=2n-1\)
Also 3, 5, 7, ....... forms an A.P. with common difference 2.
\(\therefore\) \(n^{th}\) term \(=a_n=a+(n-1)d\)
\(=3+(n-1)2\)
\(=2n-2+3\)
\(=2n+1\)
\(\therefore\) \(n^{th}\) term of series 1.3 + 3.5 + 5.7 + ……….. is \((2n-1)(2n+1)=(2n)^2-1^2=4n^2 - 1\)
Hence, \(n^{th}\) term of the series 1.3 + 3.5 + 5.7 + ……….. is \(4n^2 - 1.\)
