A gentleman decided to treat a few children in the following manner. He gives half of his total stock of toffees and one extra to the first child, and then the half of the remaining stock along with one extra to the second and continues giving away in this fashion. His total stock exhausts after he takes care of $5$ children. How many toffees were there in his stock initially?

Let us solve this by tracking how the toffees are distributed to each child.

Let us say the gentleman initially has $x$ toffees.

The pattern is: Give half of current stock + 1 extra to each child.

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First Child:

Gets: $\frac{x}{2} + 1$ toffees

Remaining stock = $x - \left[\frac{x}{2} + 1\right] = \frac{x}{2} - 1$

Second Child:

Gets: Half of remaining + 1 = $\frac{1}{2}\left(\frac{x}{2} - 1\right) + 1 = \frac{x}{4} - \frac{1}{2} + 1 = \frac{x}{4} + \frac{1}{2}$

Remaining stock = $\left(\frac{x}{2} - 1\right) - \left(\frac{x}{4} + \frac{1}{2}\right) = \frac{x}{4} - \frac{3}{2}$

Third Child:

Gets: $\frac{1}{2}\left(\frac{x}{4} - \frac{3}{2}\right) + 1 = \frac{x}{8} - \frac{3}{4} + 1 = \frac{x}{8} + \frac{1}{4}$

Remaining stock = $\left(\frac{x}{4} - \frac{3}{2}\right) - \left(\frac{x}{8} + \frac{1}{4}\right) = \frac{x}{8} - \frac{7}{4}$

Fourth Child:

Gets: $\frac{1}{2}\left(\frac{x}{8} - \frac{7}{4}\right) + 1 = \frac{x}{16} - \frac{7}{8} + 1 = \frac{x}{16} + \frac{1}{8}$

Remaining stock = $\left(\frac{x}{8} - \frac{7}{4}\right) - \left(\frac{x}{16} + \frac{1}{8}\right) = \frac{x}{16} - \frac{15}{8}$

Fifth Child:

Gets: $\frac{1}{2}\left(\frac{x}{16} - \frac{15}{8}\right) + 1 = \frac{x}{32} - \frac{15}{16} + 1 = \frac{x}{32} + \frac{1}{16}$

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Notice how each child gets:

Child 1: $\frac{x}{2} + 1$

Child 2: $\frac{x}{4} + \frac{1}{2}$

Child 3: $\frac{x}{8} + \frac{1}{4}$

Child 4: $\frac{x}{16} + \frac{1}{8}$

Child 5: $\frac{x}{32} + \frac{1}{16}$

The x-coefficient gets halved each time, and the constant term also gets halved!

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Since all toffees are exhausted after 5 children:

$\left(\frac{x}{2} + 1\right) + \left(\frac{x}{4} + \frac{1}{2}\right) + \left(\frac{x}{8} + \frac{1}{4}\right) + \left(\frac{x}{16} + \frac{1}{8}\right) + \left(\frac{x}{32} + \frac{1}{16}\right) = x$

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Let us group the x-terms and constant terms:

x-terms: $\frac{x}{2} + \frac{x}{4} + \frac{x}{8} + \frac{x}{16} + \frac{x}{32}$

Constant terms: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}$

Finding the sum of x-terms:

To add fractions, let us use common denominator 32:

$\frac{x}{2} = \frac{16x}{32}$

$\frac{x}{4} = \frac{8x}{32}$

$\frac{x}{8} = \frac{4x}{32}$

$\frac{x}{16} = \frac{2x}{32}$

$\frac{x}{32} = \frac{x}{32}$

Sum = $\frac{16x + 8x + 4x + 2x + x}{32} = \frac{31x}{32}$

Finding the sum of constant terms:

Using common denominator 16:

$1 = \frac{16}{16}$

$\frac{1}{2} = \frac{8}{16}$

$\frac{1}{4} = \frac{4}{16}$

$\frac{1}{8} = \frac{2}{16}$

$\frac{1}{16} = \frac{1}{16}$

Sum = $\frac{16 + 8 + 4 + 2 + 1}{16} = \frac{31}{16}$

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Our equation becomes:

$\frac{31x}{32} + \frac{31}{16} = x$

$\frac{31}{16} = x - \frac{31x}{32}$

$\frac{31}{16} = \frac{32x - 31x}{32}$

$\frac{31}{16} = \frac{x}{32}$

Therefore: $x = \frac{31}{16} \times 32 = 31 \times 2 = 62$

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The gentleman initially had 62 toffees in his stock.