+1072 The sum of the first $n$ terms of a certain sequence is $n(n + 1)(n + 2).$ Find the tenth term of the sequence.
+26367 The sum of the first \(n\) terms of a certain sequence is
\(n(n + 1)(n + 2)\).
Find the tenth term of the sequence.
\(\begin{array}{|c|l|l|l|l|} \hline n & s_n=n(n+1)(n+2) & \\ \hline 1 & s_1=1*2*3=6 & s_1 = a_1 & 6= a_1 & a_1 = 6 \\ \hline 2 & s_2=2*3*4=24 & s_2 = s_1+a_2 & 24=6+a_2 & a_2 = 18 \\ \hline 3 & s_3=3*4*5=60 & s_3 = s_2+a_3 & 60=24+a_3 & a_3 = 36 \\ \hline 4 & s_4=4*5*6=120 & s_4 = s_3+a_4 & 120=60+a_4 & a_4 = 60 \\ \hline 5 & s_5=5*6*7=210 & s_5 = s_4+a_5 & 210=120+a_5 & a_5 = 90 \\ \hline 6 & s_6=6*7*8=336 & s_6 = s_5+a_6 & 336=210+a_6 & a_6 = 126 \\ \hline 7 & s_7=7*8*9=504 & s_7 = s_6+a_7 & 504=336+a_7 & a_7 = 168 \\ \hline 8 & s_8=8*9*10=720 & s_8 = s_7+a_8 & 720=504+a_8 & a_8 = 216 \\ \hline 9 & s_9=9*10*11=990 & s_9 = s_8+a_9 & 990=720+a_9 & a_9 = 270 \\ \hline 10 & s_{10}=10*11*12=1320 & s_{10} = s_9+a_{10} & 1320=990+a_{10} & \mathbf{a_{10} = 330} \\ \hline \end{array}\)
+1072 I just found a way easier way, plug in 9 for n and then that would be the sum of the first nine terms of the sequence, so then you subtract that from when n=10, and then you get the tenth term: 10(10+1)(10+2)-9(9+1)(9+2)=330
