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Which recursive formula can be used to define this sequence for n > 1 ?\newline1,12,23,34,45,56, 1, 12, 23, 34, 45, 56, \ldots \newlineChoices:\newlinean=176an1 a_n = \frac{17}{6}a_{n - 1} \newlinean=an1+an2+11 a_n = a_{n - 1} + a_{n - 2} + 11 \newlinean=an111 a_n = a_{n - 1} - 11 \newlinean=an1+11 a_n = a_{n - 1} + 11

Full solution

Q. Which recursive formula can be used to define this sequence for n>1 n > 1 ?\newline1,12,23,34,45,56, 1, 12, 23, 34, 45, 56, \ldots \newlineChoices:\newlinean=176an1 a_n = \frac{17}{6}a_{n - 1} \newlinean=an1+an2+11 a_n = a_{n - 1} + a_{n - 2} + 11 \newlinean=an111 a_n = a_{n - 1} - 11 \newlinean=an1+11 a_n = a_{n - 1} + 11
  1. Identify Pattern in Sequence: To find the recursive formula, we need to look at the pattern in the sequence. Let's examine the difference between consecutive terms.\newline121=1112 - 1 = 11\newline2312=1123 - 12 = 11\newline3423=1134 - 23 = 11\newline4534=1145 - 34 = 11\newline5645=1156 - 45 = 11\newlineWe can see that each term is 1111 more than the previous term.
  2. Express nnth Term: Based on the pattern, we can express the nnth term (ana_n) as the sum of the (n1n-1)th term (an1a_{n-1}) and 1111. This gives us the recursive formula:\newlinean=an1+11a_n = a_{n-1} + 11
  3. Check Given Choices: Now let's check the given choices to see which one matches our formula:\newlineChoices: \newline[a] an=176an1a_n = \frac{17}{6}a_{n - 1} (This is not correct as it does not represent the pattern we found.)\newline[b] an=an1+an2+11a_n = a_{n - 1} + a_{n - 2} + 11 (This is not correct as it involves two previous terms and does not represent the simple addition of 1111 to the previous term.)\newline[c] an=an111a_n = a_{n - 1} - 11 (This is not correct as it subtracts 1111 instead of adding it.)\newline[d] an=an1+11a_n = a_{n - 1} + 11 (This is correct as it matches the pattern we found.)

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