Question

PLEASE HELP WITH QUESTION!!!!!


Find the sum of the series:

Answer 1

The sum of the given series
`\sum _(a=1)^(10)\:\left(2a+2\right)` is 130.

Explanation:

The sum of series need to be found is
`\sum _(a=1)^(10)\:\left(2a+2\right)` .

Apply sum rule,

`\sum x_n+y_n=\sum x_n+\sum y_n`

`\sum _(a=1)^(10)\:\left(2a+2\right).` =
`\sum 2a+\sum 2`.

Apply the constant multiplication rule,

`\sum c\cdot a_n=c\cdot \sum a_n` .

`\sum 2\cdot a=2\cdot \sum a`.

`\sum _(a=1)^(10)n` = 1+2+3+4+5+6+7+8+9+10.

`\sum _(a=1)^(10)n` = 55.

`\sum 2\cdot a`=2×55.

`\sum 2\cdot a`=110.

To find
`\sum _(a=1)^(10)2`,

Apply sum rule,
`\sum _(k=1)^n\:a\:=\:a\cdot n` ,

`\sum _(a=1)^(10)2` =2×10.

`\sum _(a=1)^(10)2` = 20.

`\sum _(a=1)^(10)\:\left(2a+2\right)` = 110+20.

`\sum _(a=1)^(10)\:\left(2a+2\right)` =130.