Question

Find the value of the sum….Use a summation formula, if possible.

Answer 1

we have the expression

`\sum ^(29)_(j\mathop=1)(5j^2-5j+3)`

Applying property of the summation formula

`\sum ^(29)_{j\mathop{=}1}(5j^2-5j+3)=5\sum ^(29)_{j\mathop{=}1}j^2-5\sum ^(29)_{j\mathop{=}1}j+\sum ^(29)_{j\mathop{=}1}3`

step 1

Find out

`5\sum ^(29)_{j\mathop{=}1}j^2`

Applying formulas

`5\sum ^(29)_{j\mathop{=}1}j^2=5\cdot\lbrack(n(n+1)(2n+1))/(6)\rbrack`

Substitute the value of n=29

`5\sum ^(29)_{j\mathop{=}1}j^2=5\cdot\lbrack(29(29+1)(2(29)+1))/(6)\rbrack=42,775`

step 2

Find out

`5\sum ^(29)_{j\mathop{=}1}j`

Applying formulas

`5\sum ^(29)_{j\mathop{=}1}j=5\cdot\lbrack(n(n+1))/(2)\rbrack`

For n=29

substitute

`5\sum ^(29)_{j\mathop{=}1}j=5\cdot\lbrack(29(29+1))/(2)\rbrack=2,175`

step 3

Find out

`\sum ^(29)_{j\mathop{=}1}3`

applying formulas

`\sum ^(29)_{j\mathop{=}1}3=3\cdot n=3\cdot29=87`

therefore

`\sum ^(29)_{j\mathop{=}1}(5j^2-5j+3)=42,775-2,175+87=40,687`

the answer is 40,687