Question

Evaluate the sum summation from k equals 1 to 9 k left parenthesis 5 k plus 2 right parenthesis.

Answer 1

We need to evaluate:

`\sum_(k=1)^(9)(5k+2)` Where k=1,2,3,4,5..........9

When k=1 then 5k+2= 5(1)+2=5+2=7

Similarly when k=2 then 5k+2=5(2)+2=10+2=12

When k=3 then 5k+2=5(3)+2=15+2=17

When k=4 then 5k+2=5(4)+2=20+2=22

When k=5 then 5k+2=5(5)+2=25+2=27

When k=6 then 5k+2=5(6)+2=30+2=32

When k=7 then 5k+2=5(7)+2=35+2=37

When k=8 then 5k+2=5(8)+2=40+2=42

When k=9 then 5k+2=5(9)+2=45+2=47.

Now to find
`\sum_(k=1)^(9)(5k+2)` add all the numbers. So,

`\sum_(k=1)^(9)(5k+2)`=7+12+17+22+27+32+37+42+47 =243.

So,
`\sum_(k=1)^(9)(5k+2)`=243