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Find S10 for 2 + 5 + 8 +... @mathstudent55 i got 270, is that right?
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\(\sum_{n=1}^{10}(2+3(n-1))=\sum_{n=1}^{10}-1+3n=\sum_{n=1}^{10}-1+3\sum_{n=1}^{10}n=-10+3\dfrac{10*11}{2}=\\-10+3*5*11=165-10=155\)
\(a_1 = 2\), \(a_2 = 5\), and \(a_3 = 8\) We are dealing with an arithmetic series. \(d = a_2 - a_1 = 5 - 2 = 3\) \(a_n = a_1 + (n - 1)d\) \(a_{10} = 2 + (10 - 1)3 = 2 + 9(3) = 29\) \(S_n = \dfrac{n(a_1 + a_n)}{2} \) \(S_n = \dfrac{10(2 + 29)}{2} = 155\)
that seems... right...
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