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_{n}= 3n

^{2}- n. Show that the nth term of the sequence is given by a

_{n}= 6n - 4.

so far i have done:

S

_{n}= (n / 2) (a

_{1}+ a

_{n}) = 3n

^{2}- n

i solved for a

_{1}+ a

_{n}= 6n - 2

i also have a

_{n}= a

_{1}+ d(n-1).

i don't know what do to next. please help me.