Question

Complete the series: \(3, 7, 15, 31, 63, \ldots\)

• A. \(95\)
• B. \(111\)
• C. \(127\)
• D. \(128\)

Hint:

Check if each term relates to a power of 2, such as 2 raised to an increasing power minus 1, in addition to checking the relation between consecutive terms. Spotting the exponential form usually gives the next term faster than repeating the same calculation each time.

Answer 1

The Correct Option is C

Concept: Many number series follow a recursive pattern where each term is obtained by performing an operation on the previous term. A common pattern is multiplying the previous term and then adding a constant.

Step 1: Observe the pattern between consecutive terms. \[ 3 \rightarrow 7 \rightarrow 15 \rightarrow 31 \rightarrow 63 \] Check the relation: \[ 3 \times 2 + 1 = 7 \] \[ 7 \times 2 + 1 = 15 \] \[ 15 \times 2 + 1 = 31 \] \[ 31 \times 2 + 1 = 63 \] Thus, the pattern is: \[ a_{n+1} = 2a_n + 1 \]

Step 2: Apply the pattern to find the next term. \[ 63 \times 2 + 1 = 126 + 1 \] \[ = 127 \]

Answer 2

Concept:

• A recursive pattern such as double and add 1 can often be rewritten as a direct exponential formula in terms of the position n, which finds any term without repeating the recursion from the start.
• Here each term is one less than a power of 2: $a_n = 2^{n+1} - 1$.

Step 1: Write each given term as a power of 2 minus 1.
$3 = 2^2 - 1$, $\ 7 = 2^3 - 1$, $\ 15 = 2^4 - 1$

Step 2: Confirm the pattern against the remaining terms.
$31 = 2^5 - 1$, $\ 63 = 2^6 - 1$
Every term matches the formula $a_n = 2^{n+1} - 1$ for $n = 1, 2, 3, 4, 5$.

Step 3: Apply the formula for the sixth term, $n = 6$.
$a_6 = 2^{7} - 1 = 128 - 1$

Final Answer: $127$