find the value of 7.8 bar +7.88 bar +7.888 bar + 7.8888 bar + ….. up to 2014 times (interns of non terminating recurring decimal form
The Correct Answer and Explanation is:
The given problem involves summing a sequence of numbers, each with a recurring decimal. The sequence is:7.8‾+7.888‾+7.8888‾+7.88888‾+…7.\overline{8} + 7.88\overline{8} + 7.888\overline{8} + 7.8888\overline{8} + \dots7.8+7.888+7.8888+7.88888+…
The sequence continues with each successive term having one more “8” in the decimal places and the number of terms is 2014.
Step 1: Converting each term into a non-terminating recurring decimal
Let’s express each term in the sequence as a non-terminating recurring decimal:
- 7.8‾=7+0.8‾7.\overline{8} = 7 + 0.\overline{8}7.8=7+0.8, which is a repeating decimal.
- 7.888‾=7+0.888‾7.88\overline{8} = 7 + 0.88\overline{8}7.888=7+0.888.
- 7.8888‾=7+0.8888‾7.888\overline{8} = 7 + 0.888\overline{8}7.8888=7+0.8888, and so on.
Each term can be written in the form:7+89, 7+890, 7+8900,…7 + \frac{8}{9}, \, 7 + \frac{8}{90}, \, 7 + \frac{8}{900}, \dots7+98,7+908,7+9008,…
The repeating part 8‾\overline{8}8 in decimal corresponds to the fraction 89\frac{8}{9}98, but the denominator changes depending on how many digits are repeated after the decimal point.
Step 2: Recognizing the general formula for each term
The general form for the nnn-th term in the sequence is:7+810n−17 + \frac{8}{10^n – 1}7+10n−18
Where nnn is the number of digits after the decimal. For the first term, n=1n = 1n=1, for the second term n=2n = 2n=2, and so on up to n=2014n = 2014n=2014.
Step 3: Summing the sequence
The sum can be written as:S=∑n=12014(7+810n−1)S = \sum_{n=1}^{2014} \left( 7 + \frac{8}{10^n – 1} \right)S=n=1∑2014(7+10n−18)
This sum can be split into two parts:
- The constant part: ∑n=120147=7×2014=14098\sum_{n=1}^{2014} 7 = 7 \times 2014 = 14098∑n=120147=7×2014=14098
- The fractional part: ∑n=12014810n−1\sum_{n=1}^{2014} \frac{8}{10^n – 1}∑n=1201410n−18
Step 4: Evaluating the fractional sum
The sum of the fractional parts involves a series of terms where each term decreases as nnn increases. This series converges rapidly, and we can approximate the sum of the fractions. For large nnn, the term 810n−1\frac{8}{10^n – 1}10n−18 becomes very small.
Thus, the total sum is:S≈14098+∑n=12014810n−1S \approx 14098 + \sum_{n=1}^{2014} \frac{8}{10^n – 1}S≈14098+n=1∑201410n−18
The sum of the fractions ∑n=12014810n−1\sum_{n=1}^{2014} \frac{8}{10^n – 1}∑n=1201410n−18 is roughly equal to:∑n=12014810n≈0.8888\sum_{n=1}^{2014} \frac{8}{10^n} \approx 0.8888n=1∑201410n8≈0.8888
So, the total sum SSS is approximately:S≈14098+0.8888=14098.8888S \approx 14098 + 0.8888 = 14098.8888S≈14098+0.8888=14098.8888
Final Answer:
S≈14098.8888S \approx 14098.8888S≈14098.8888
Thus, the value of the series up to 2014 terms is approximately 14098.8888.
