Binson's answer to Carol's Secondary 1 Maths Singapore question.
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To find the sum of the first three terms of the sequence with the general term Tⁿ = 1/(2n+1), we can substitute the values of n = 1, 2, and 3 into the general term and then add them together.
T¹ = 1/(21+1) = 1/3
T² = 1/(22+1) = 1/5
T³ = 1/(2*3+1) = 1/7
Sum of the first three terms = T¹ + T² + T³ = 1/3 + 1/5 + 1/7
To add these fractions, we need to find a common denominator, which is the least common multiple (LCM) of 3, 5, and 7. The LCM of these three numbers is 105.
Converting the fractions to have a common denominator of 105:
1/3 = (1/3) * (35/35) = 35/105
1/5 = (1/5) * (21/21) = 21/105
1/7 = (1/7) * (15/15) = 15/105
Sum of the first three terms = (35/105) + (21/105) + (15/105) = 71/105
Therefore, the sum of the first three terms of the sequence is 71/105.
T¹ = 1/(21+1) = 1/3
T² = 1/(22+1) = 1/5
T³ = 1/(2*3+1) = 1/7
Sum of the first three terms = T¹ + T² + T³ = 1/3 + 1/5 + 1/7
To add these fractions, we need to find a common denominator, which is the least common multiple (LCM) of 3, 5, and 7. The LCM of these three numbers is 105.
Converting the fractions to have a common denominator of 105:
1/3 = (1/3) * (35/35) = 35/105
1/5 = (1/5) * (21/21) = 21/105
1/7 = (1/7) * (15/15) = 15/105
Sum of the first three terms = (35/105) + (21/105) + (15/105) = 71/105
Therefore, the sum of the first three terms of the sequence is 71/105.
Date Posted:
1 year ago
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