Question

Find the 1st term, last term and the sum for the finite arithmetic series.

Answer 1

Given that,

`\sum ^(30)_(n\mathop=2)(3n-1)`

Simplifying we get,

`\sum ^(30)_{n\mathop{=}2}(3n-1)=\sum ^(30)_{n\mathop{=}2}3n+\sum ^(30)_{n\mathop{=}2}1`
`=3\sum ^(30)_{n\mathop{=}2}n+\sum ^(30)_{n\mathop{=}2}1`

we have that,

`\sum ^n_(n\mathop=1)1=n`

If n is from 2 to n we get,

`\sum ^n_{n\mathop{=}2}1=n-1`

Also,

`\sum ^k_(n\mathop=1)n=(k(k+1))/(2)`

If n is from 2 to n we get,

`\sum ^k_(n\mathop=2)n=(k(k+1))/(2)-1`

Using this and substituting in the required expression we get,

`=3\lbrack(30*31)/(2)-1\rbrack+30-1`
`=3(464)+29`
`=1421`

Answer is: 1421