Question

find the sum to n terms of the series whose n terms is 3n^2+4n-2.

Answer 1

$\huge \bf{S = \frac{n(n + 1)}{2}} \\ \\ \sf > S = \frac{ {3n}^{2} + 4n - 2( {3n}^{2} + 4n - 2 + 1)}{2} \\ \\ \tiny{ > \sf S = \frac{{3n}^{2} + 4n - 2({3n}^{2} + 4n - 1)}{2}} \\ \\ \sf \tiny{ > S = \frac{[ {3n}^{2}({3n}^{2} + 4n - 1) + 4n({3n}^{2} + 4n - 1) - 2({3n}^{2} + 4n - 1)]}{2}} \\ \\ \tiny{\sf > S = \frac{ {9n}^{4} + {12n}^{3} - {3n}^{2} + {12n}^{3} + {16n}^{2} - 4n - {6n}^{2} - 8n + 2 }{2} } \\ \\ \tiny{ \sf >S = \frac{ {9n}^{4} + {24n}^{3} + {7n}^{2} - 12n + 2 }{2} } \\ \\ \tiny{ \sf >S = \frac{ \cancel2( {4.5n}^{4} + {12n}^{3} + 3.5 {n}^{2} - 6n + 1 )}{ \cancel2} }$

Hence⤵

Required Sum is $\tt{{4.5n}^{4} + {12n}^{3} + 3.5 {n}^{2} - 6n + 1}$.