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{[b(1)=16],[b(n)=b(n-1)+1]:} Find the 2^("nd ") term in the sequence.

{b(1)=16b(n)=b(n−1)+1 \left\{\begin{array}{l} b(1)=16 \\ b(n)=b(n-1)+1 \end{array}\right. \newlineFind the 2nd  2^{\text {nd }} term in the sequence.

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Full solution

Q. {b(1)=16b(n)=b(n−1)+1 \left\{\begin{array}{l} b(1)=16 \\ b(n)=b(n-1)+1 \end{array}\right. \newlineFind the 2nd  2^{\text {nd }} term in the sequence.
  1. Given first term: We are given the first term of the sequence: b(1)=16b(1) = 16. The rule for the sequence is that each term is one more than the previous term, which is defined by b(n)=b(n−1)+1b(n) = b(n-1) + 1.
  2. Apply rule for second term: To find the second term, we apply the rule to the first term. Since the second term is b(2)b(2), we use n=2n=2 in the formula: b(2)=b(2−1)+1=b(1)+1b(2) = b(2-1) + 1 = b(1) + 1.
  3. Substitute and calculate: We know that b(1)=16b(1) = 16, so we substitute this value into the equation: b(2)=16+1b(2) = 16 + 1.
  4. Final result: Perform the calculation: b(2)=17b(2) = 17.

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