Answer:
Given that,
[tex]\sum ^{30}_{n\mathop=2}(3n-1)[/tex]
Simplifying we get,
[tex]\sum ^{30}_{n\mathop{=}2}(3n-1)=\sum ^{30}_{n\mathop{=}2}3n+\sum ^{30}_{n\mathop{=}2}1[/tex][tex]=3\sum ^{30}_{n\mathop{=}2}n+\sum ^{30}_{n\mathop{=}2}1[/tex]
we have that,
[tex]\sum ^n_{n\mathop=1}1=n[/tex]
If n is from 2 to n we get,
[tex]\sum ^n_{n\mathop{=}2}1=n-1[/tex]
Also,
[tex]\sum ^k_{n\mathop=1}n=\frac{k(k+1)}{2}[/tex]
If n is from 2 to n we get,
[tex]\sum ^k_{n\mathop=2}n=\frac{k(k+1)}{2}-1[/tex]
Using this and substituting in the required expression we get,
[tex]=3\lbrack\frac{30\times31}{2}-1\rbrack+30-1[/tex][tex]=3(464)+29[/tex][tex]=1421[/tex]
Answer is: 1421
