Find the sum of the finite series. sum_(i=1)^(50)(2i-1)(2i+1) ____

Simplify Expression: We start by simplifying the expression (2i-1)(2i+1) using the difference of squares formula, which states that a^2 - b^2 = (a - b)(a + b). Here, a = 2i and b = 1, so the expression simplifies to 4i^2 - 1.Distribute Summation: Next, we distribute the summation from i=1 to 50 into 4i^2 - 1. This separates the series into two parts: sum_(i=1)^(50)4i^2 and -sum_(i=1)^(50)1.Evaluate First Term: We evaluate the first term sum_(i=1)^(50)4i^2 using the summation formula for the sum of squares, which is sum_(i=1)^(n)i^2 = (n(n+1)(2n+1))/6. Plugging in n = 50, we get: sum_(i=1)^(50)4i^2 = 4 * ((50(50+1)(2*50+1))/6) = 4 * ((50*51*101)/6) = 4 * (2550*101)/6 = 4 * (257550/6) = 4 * 42925 = 171700Evaluate Second Term: We evaluate the second term sum_(i=1)^(50)1 which is simply the sum of 1 repeated 50 times. This is equal to 50.Subtract and Find Answer: Finally, we subtract the result of the second term from the first term to get the final answer: 171700 - 50 = 171650.

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Find the sum of the finite series. $\newline$ $\sum_{i=1}^{50} (2i-1)(2i+1)$ $\newline$ ______

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Algebra 2

Sum of finite series starts from 1

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Q. Find the sum of the finite series.

\newline

\sum_{i=1}^{50} (2i-1)(2i+1)

\newline

______

Simplify Expression: We start by simplifying the expression $(2i-1)(2i+1)$ using the difference of squares formula, which states that $a^2 - b^2 = (a - b)(a + b)$ . Here, $a = 2i$ and $b = 1$ , so the expression simplifies to $4i^2 - 1$ .
Distribute Summation: Next, we distribute the summation from $i=1$ to $50$ into $4i^2 - 1$ . This separates the series into two parts: $\sum_{i=1}^{50}4i^2$ and $-\sum_{i=1}^{50}1$ .
Evaluate First Term: We evaluate the first term $\sum_{i=1}^{50}4i^2$ using the summation formula for the sum of squares, which is $\sum_{i=1}^{n}i^2 = \frac{n(n+1)(2n+1)}{6}$ . Plugging in $n = 50$ , we get: $\newline$ $\sum_{i=1}^{50}4i^2 = 4 \times \left(\frac{50(50+1)(2\times50+1)}{6}\right)$ $\newline$ $= 4 \times \left(\frac{50\times51\times101}{6}\right)$ $\newline$ $= 4 \times \frac{2550\times101}{6}$ $\newline$ $= 4 \times \frac{257550}{6}$ $\newline$ $= 4 \times 42925$ $\newline$ $= 171700$
Evaluate Second Term: We evaluate the second term $\sum_{i=1}^{50}1$ which is simply the sum of $1$ repeated $50$ times. This is equal to $50$ .
Subtract and Find Answer: Finally, we subtract the result of the second term from the first term to get the final answer: $171700 - 50 = 171650$ .