Find the sum of the first 50 terms in this geometric series: 1+(10)/(11)+(100)/(121)dots Choose 1 answer: (A) 0.01 (B) 0.99 (C) 10.91 (D) 11

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Find the sum of the first 50 terms in this geometric series:  1+(10)/(11)+(100)/(121)dots Choose 1 answer: (A) 0.01 (B) 0.99 (C) 10.91 (D) 11

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Identifying the series: We are given a geometric series: 1 + (10/11) + (100/121) + ... First, we need to identify the first term (a) and the common ratio (r) of the series. The first term a is clearly 1. To find the common ratio r, we divide the second term by the first term: r = (10/11) / 1 = 10/11Finding the common ratio: Now that we have the first term a = 1 and the common ratio r = 10/11, we can use the formula for the sum of the first n terms of a geometric series: S_n = a(1 - r^n) / (1 - r), where S_n is the sum of the first n terms. We want to find the sum of the first 50 terms, so n = 50.Using the formula for the sum: Let's plug the values into the formula: S_50 = 1(1 - (10/11)^50) / (1 - 10/11)Plugging in the values: Simplify the denominator: 1 - 10/11 = 11/11 - 10/11 = 1/11 So, S_50 = 1(1 - (10/11)^50) / (1/11)Simplifying the denominator: Now, we multiply both the numerator and the denominator by 11 to simplify the fraction: S_50 = 11(1 - (10/11)^50)Multiplying by 11: We can now calculate (10/11)^50, but since this is a very small number (because 10/11 is less than 1 and we are raising it to the 50th power), it will have a negligible effect on the sum. Therefore, we can approximate the sum by ignoring this term: S_50 ≈ 11(1 - 0) S_50 ≈ 11Approximating the sum: The sum of the first 50 terms in the given geometric series is approximately 11.

Find the sum of the first $50$ terms in this geometric series: $\newline$ $1+\frac{10}{11}+\frac{100}{121} \ldots$ $\newline$ Choose $1$ answer: $\newline$ (A) $0$ . $01$ $\newline$ (B) $0$ . $99$ $\newline$ (C) $\mathbf{1 0 . 9 1}$ $\newline$ (D) $11$

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Find the sum of the first 505050 terms in this geometric series:\newline1+1011+100121… 1+\frac{10}{11}+\frac{100}{121} \ldots 1+1110​+121100​…\newlineChoose 111 answer:\newline(A) 000.010101\newline(B) 000.999999\newline(C) 10.91 \mathbf{1 0 . 9 1} 10.91\newline(D) 111111

Find the sum of the first $50$ terms in this geometric series: $\newline$ $1+\frac{10}{11}+\frac{100}{121} \ldots$ $\newline$ Choose $1$ answer: $\newline$ (A) $0$ . $01$ $\newline$ (B) $0$ . $99$ $\newline$ (C) $\mathbf{1 0 . 9 1}$ $\newline$ (D) $11$