Evaluate -sum_(i=1)^(100)(-1)^(i)((5!×5)/(4!))^((1)/(2))

Simplify Expression Inside Summation: We first simplify the expression inside the summation. The expression ((5!×5)/(4!))^((1)/(2)) can be simplified since 5! = 5×4! and thus the 4! in the numerator and denominator will cancel out.Calculate Simplified Expression: After canceling out 4!, we are left with (5×5)^((1)/(2)) which simplifies to 5^(1/2) * 5^(1/2) = 5.Factor Out Constant: Now we have the simplified series -sum_(i=1)^(100)(-1)^(i) * 5. Since 5 is a constant, we can factor it out of the summation.Evaluate Alternating Series: The series now becomes -5 * sum_(i=1)^(100)(-1)^(i). This is an alternating series where the sign changes with each term.Final Result: To evaluate the alternating series sum_(i=1)^(100)(-1)^(i), we notice that for odd i, (-1)^(i) is -1, and for even i, (-1)^(i) is 1. Since there are an equal number of odd and even terms (50 of each), the series will sum to 0.Multiplying the result of the series by -5, we get -5 * 0 = 0.

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Evaluate $-\sum_{i=1}^{100}(-1)^{i}\left(\frac{5!\times5}{4!}\right)^{\frac{1}{2}}$

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Algebra 2

Sum of finite series starts from 1

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Q. Evaluate

-\sum_{i=1}^{100}(-1)^{i}\left(\frac{5!\times5}{4!}\right)^{\frac{1}{2}}

Simplify Expression Inside Summation: We first simplify the expression inside the summation. The expression $\left(\frac{5!\times 5}{4!}\right)^{\frac{1}{2}}$ can be simplified since $5! = 5\times 4!$ and thus the $4!$ in the numerator and denominator will cancel out.
Calculate Simplified Expression: After canceling out $4!$ , we are left with $(5\times5)^{(1)/(2)}$ which simplifies to $5^{1/2} \times 5^{1/2} = 5$ .
Factor Out Constant: Now we have the simplified series $-\sum_{i=1}^{100}(-1)^{i} \times 5$ . Since $5$ is a constant, we can factor it out of the summation.
Evaluate Alternating Series: The series now becomes $-5 \times \sum_{i=1}^{100}(-1)^{i}$ . This is an alternating series where the sign changes with each term.
Final Result: To evaluate the alternating series $\sum_{i=1}^{100}(-1)^{i}$ , we notice that for odd $i$ , $(-1)^{i}$ is $-1$ , and for even $i$ , $(-1)^{i}$ is $1$ . Since there are an equal number of odd and even terms ( $50$ of each), the series will sum to $0$ .
Final Result: To evaluate the alternating series $\sum_{i=1}^{100}(-1)^{i}$ , we notice that for odd $i$ , $(-1)^{i}$ is $-1$ , and for even $i$ , $(-1)^{i}$ is $1$ . Since there are an equal number of odd and even terms ( $50$ of each), the series will sum to $0$ .Multiplying the result of the series by $-5$ , we get $i$ $0$ .