Sum of first n terms of 1.4 + 4.7 + 7.10 + 10.13 + ……………… is 

Correct option is (B) \(9 \sum n^2 - 3 \sum n - 2n\)

1, 4, 7, 10, ........ will form an A.P. whose \(n^{th}\) term is \(a_n=a+(n-1)d\)

\(=1+(n-1)3\)

\(=3n-3+1\)

\(=3n-2\)

Also 4, 7, 10, 13, ......... will form another A.P. whose \(n^{th}\) term is \(a_n=a+(n-1)d\)

\(=4+(n-1)3\)

\(=3n-3+n\)

\(=3n+1\)

Now, 1.4+4.7+7.10+10.13+.........+n terms

= 1.4+4.7+7.10+10.13+.........+(3n-2)(3n+1)

\(=\sum(3n-2)(3n+1)\)

\(=\sum(9n^2-6n+3n-2)\)

\(=\sum(9n^2-3n-2)\)

\(=9\sum n^2-3\sum n-2\sum1\)

\(=9\sum n^2-3\sum n-2n\)         \((\sum1=n)\)

Correct option is B) 9Σn² – 3Σn – 2n