Find the sum of the finite series. sum_(i=3)^(120)i(i-5) ____

Expand and Simplify: Simplify the expression i(i-5) by expanding it. This gives us i^2 - 5i.Distribute Summation: Distribute the summation from 3 to 120 into i^2 - 5i. This gives us sum_(i=3)^(120)i^2 - sum_(i=3)^(120)5i.Evaluate First Term: Evaluate the first term sum_(i=3)^(120)i^2 using the summation rule for squares of integers. sum_(i=1)^(n)i^2 = (n(n+1)(2n+1))/6 We need to subtract the sum of squares from 1 to 2 from the sum of squares from 1 to 120. sum_(i=3)^(120)i^2 = sum_(i=1)^(120)i^2 - sum_(i=1)^(2)i^2 = ((120(120+1)(2*120+1))/6) - ((2(2+1)(2*2+1))/6) = ((120*121*241)/6) - ((2*3*5)/6) = (2905200/6) - (30/6) = 484200 - 5 = 484195Evaluate Second Term: Evaluate the second term sum_(i=3)^(120)5i using the summation rule for arithmetic series. sum_(i=1)^(n)i = (n(n+1))/2 We need to subtract the sum from 1 to 2 from the sum from 1 to 120 and then multiply by 5. sum_(i=3)^(120)5i = 5 * (sum_(i=1)^(120)i - sum_(i=1)^(2)i) = 5 * (((120(120+1))/2) - ((2(2+1))/2)) = 5 * ((120*121/2) - (2*3/2)) = 5 * (7260 - 3) = 5 * 7257 = 36285Subtract and Final Answer: Subtract the second term from the first term to get the final answer. Final Answer = 484195 - 36285 = 447910

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Math Problems

Algebra 2

Sum of finite series not start from 1

Full solution

Q. Find the sum of the finite series.

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\sum_{i=3}^{120} i(i-5)

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Expand and Simplify: Simplify the expression $i(i-5)$ by expanding it. This gives us $i^2 - 5i$ .
Distribute Summation: Distribute the summation from $3$ to $120$ into $i^2 - 5i$ . This gives us $\sum_{i=3}^{120}i^2 - \sum_{i=3}^{120}5i$ .
Evaluate First Term: Evaluate the first term $\sum_{i=3}^{120}i^2$ using the summation rule for squares of integers. $\sum_{i=1}^{n}i^2 = \frac{n(n+1)(2n+1)}{6}$ We need to subtract the sum of squares from $1$ to $2$ from the sum of squares from $1$ to $120$ . $\sum_{i=3}^{120}i^2 = \sum_{i=1}^{120}i^2 - \sum_{i=1}^{2}i^2$ $= \left(\frac{120(120+1)(2\cdot120+1)}{6}\right) - \left(\frac{2(2+1)(2\cdot2+1)}{6}\right)$ $= \left(\frac{120\cdot121\cdot241}{6}\right) - \left(\frac{2\cdot3\cdot5}{6}\right)$ $= \left(\frac{2905200}{6}\right) - \left(\frac{30}{6}\right)$ $= 484200 - 5$ $= 484195$
Evaluate Second Term: Evaluate the second term $\sum_{i=3}^{120}5i$ using the summation rule for arithmetic series. $\newline$ $\sum_{i=1}^{n}i = \frac{n(n+1)}{2}$ $\newline$ We need to subtract the sum from $1$ to $2$ from the sum from $1$ to $120$ and then multiply by $5$ . $\newline$ $\sum_{i=3}^{120}5i = 5 \times (\sum_{i=1}^{120}i - \sum_{i=1}^{2}i)$ $\newline$ $= 5 \times \left(\frac{120(120+1)}{2} - \frac{2(2+1)}{2}\right)$ $\newline$ $= 5 \times \left(\frac{120\times121}{2} - \frac{2\times3}{2}\right)$ $\newline$ $= 5 \times (7260 - 3)$ $\newline$ $= 5 \times 7257$ $\newline$ $= 36285$
Subtract and Final Answer: Subtract the second term from the first term to get the final answer. $\newline$ Final Answer = $484195 - 36285$ $\newline$ = $447910$