A sequence of 24 positive numbers a_1, a_2, \dots, a_{24} is defined by a_n = \frac{120}{n… — GRE General Test

A sequence of 24 positive numbers $a_1, a_2, \dots, a_{24}$ is defined by $a_n = \frac{120}{n(n+1)}$ for each integer $n$ from $1$ to $24$ . Let $T = \sum_{n=1}^{24} a_n$ be the exact sum of all 24 terms. Which of the following statements must be true? Select all that apply.

The value of $T$ rounded to the nearest integer is $115$ .Cevap
The sum of the first 4 terms, $\sum_{n=1}^{4} a_n$ , represents more than $80\%$ of the total sum $T$ .Cevap
C
The sum of the first 2 terms, $a_1 + a_2$ , is equal to $100$ .
D
The percent error in estimating the total sum $T$ by using an upper bound of $120$ is exactly $4.0\%$ .
E
For the sequence term formula $a_n = f(n)$ , the value of $f(1 + 2)$ is equal to $f(1) + f(2)$ .

Cevap

The statements confirming that

T

rounded to the nearest integer is

115

, and that the sum of the first 4 terms represents more than

80\%

T

, are both correct.

The exact sum of the telescoping sequence simplifies to

T = 120 \left(1 - \frac{1}{25}\right) = 115.2

. Rounding

115.2

to the nearest integer gives

115

. Furthermore, the partial sum of the first four terms is

120 \left(1 - \frac{1}{5}\right) = 96

, which accounts for

\frac{96}{115.2} = \frac{5}{6} \approx 83.33\%

of the total sum, exceeding

80\%

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Decompose the sequence formula using partial fractions.

a_n = \frac{120}{n(n+1)} = 120 \left( \frac{1}{n} - \frac{1}{n+1} \right)

Rewriting the terms as partial fractions converts the sum into a telescoping series.

Calculate the exact total sum

T

T = 120 \left[ \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{24} - \frac{1}{25}\right) \right] = 120 \left(1 - \frac{1}{25}\right) = 120 \times 0.96 = 115.2

All intermediate terms cancel out, leaving only the first and last components.

Evaluate the statement regarding rounding

T

to the nearest integer.

115.2

rounded to the nearest integer is

115

, making the rounding statement true.

Since the decimal part

.2

is less than

.5

, the number rounds down to

115

Evaluate the sum of the first 4 terms and compare its percentage to

T

\sum_{n=1}^{4} a_n = 120 \left(1 - \frac{1}{5}\right) = 96

. The percentage is

\frac{96}{115.2} = \frac{5}{6} \approx 83.33\% > 80\%

Comparing

83.33\%

80\%

confirms that the partial sum statement is true.

Anahtar Kavram

Telescoping Series Summation and Percent Estimation

Soru

Cevap

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Anahtar Kavram

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