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

Question

asked May 15, 2023 138k views

1 Answer

Anass Kartit · Answer 1 · 2023-05-19T07:24:38+0000

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

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