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

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

Substitute the value of n=29

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

step 2

Find out

$5\sum ^{29}_{j\mathop{=}1}j$

Applying formulas

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

For n=29

substitute

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

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