Couple of ways to answer this ....

Method 1: Find the patternLet's say A(1) = x for a while

Now, let's write the series bearing in mind that each term is 1 less than twice its preceding term.

\(A1 = x\)

\(A2 = 2x -1\)

\(A3 = 4x -3\)

\(A4 = 8x -7\)

\(A5 = 16x -15\)

\(A6 = 32x -31\)

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Notice the following

\(A2 - A1 = x - 1\)

\(A3 - A2 = 2(x-1) = 2^1(x-1)\)

\(A4 - A3 = 4(x-1) = 2^2(x-1)\)

\(A5 - A4 = 8(x-1) = 2^3(x-1)\)

\(A6 - A5 = 16(x-1) = 2^4(x-1)\)

\(=> A24 - A23 = 2^{22}(x-1)\)

\(=> A24 - A23 = 2^{22}(3-1) = 2^{23}\) (Given x = 3 in the question)

\(=> A24 - A23 = 2^{23}\)

Method 2: Going to the start of the seriesGiven \(A(n+1) = 2A(n) - 1\)

\(=> A(n+1) + 1 = 2A(n)\) --------(i)

Bearing this format in mind, let us write the given question the following way

\(A(24) - A(23)\)

\(= A(24) + 1 - A(23) - 1\)

\(= [A(24) +1 ] - [A(23) + 1]\)

\(= 2A(23) - 2A(22)\) ( From (i) )

\(= 2[A(23) - A(22)]\)

Similarly \(A(23) - A(22)\) will equal \(2[A(22) - A(21)]\)

\(\therefore A(24) - A(23) = 2^2[A(22) - A(21)]\)

Continuing this, we get

\(A(24) - A(23) = 2^{22}[A2 - A1]\)

\(\therefore A(24) - A(23) = 2^{22}[5 - 3 ] = 2^{23}\)

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