Calculating the Value of 9p² + p for a Repeating Digit Number

The question asks us to find the value of the expression $9p^2 + p$ where $p$ is a number consisting of $n$ digits, all equal to 1.

Let's first understand the structure of the number $p$. If $p$ has $n$ digits, and all digits are 1, we can write it as:

$\qquad p = 111...1$ ($n$ times)

This number can be expressed mathematically. For example:

• If $n=1$, $p=1$.
• If $n=2$, $p=11$.
• If $n=3$, $p=111$.

In general, a number consisting of $n$ ones can be represented as the sum of powers of 10:

$\qquad p = 10^{n-1} + 10^{n-2} + ... + 10^1 + 10^0$

This is a geometric series with first term $a = 10^0 = 1$, common ratio $r = 10$, and $n$ terms. The sum of a geometric series is given by $S_n = a \frac{r^n - 1}{r - 1}$.

Using this formula, we can write $p$ as:

$\qquad p = 1 \cdot \frac{10^n - 1}{10 - 1} = \frac{10^n - 1}{9}$

Now we need to find the value of $9p^2 + p$. We will substitute the expression for $p$ into this expression:

$\qquad 9p^2 + p = 9 \left(\frac{10^n - 1}{9}\right)^2 + \frac{10^n - 1}{9}$

Let's simplify this expression step by step:

$\qquad 9 \left(\frac{10^n - 1}{9}\right)^2 = 9 \cdot \frac{(10^n - 1)^2}{9^2} = 9 \cdot \frac{(10^n - 1)^2}{81} = \frac{(10^n - 1)^2}{9}$

So the expression becomes:

$\qquad 9p^2 + p = \frac{(10^n - 1)^2}{9} + \frac{10^n - 1}{9}$

We can factor out the common term $\frac{10^n - 1}{9}$:

$\qquad 9p^2 + p = \frac{10^n - 1}{9} \left[ (10^n - 1) + 1 \right]$

Simplify the term inside the square brackets:

$\qquad (10^n - 1) + 1 = 10^n - 1 + 1 = 10^n$

Substitute this back into the expression:

$\qquad 9p^2 + p = \frac{10^n - 1}{9} \cdot 10^n$

Recall that $p = \frac{10^n - 1}{9}$. Substitute $p$ back into the expression:

$\qquad 9p^2 + p = p \cdot 10^n$

This can also be written as $10^n p$. Let's compare this result with the given options.

The options are:

1. $10^n p$
2. $2p \cdot 10^n$
3. $10^n p - 1$
4. $10^n p + 1$

Our calculated value, $10^n p$, matches the first option.

Therefore, the value of $9p^2 + p$ is $10^n p$ when $p$ is a number consisting of $n$ digits, all equal to 1.

Revision Table: Key Concepts

Additional Information: Repeating Digit Numbers

Numbers formed by repeating the same digit are interesting. A number with $n$ digits, all equal to a digit $d$, can be represented. For example, if the digit is 3 and there are $n$ digits, the number is $333...3$ ($n$ times).

This number can be written as $3 \times 111...1$ ($n$ times). Using our previous result for $p$, this number is $3 \times \frac{10^n - 1}{9} = \frac{3(10^n - 1)}{9} = \frac{10^n - 1}{3}$.

In general, a number with $n$ digits, all equal to digit $d$, can be written as $d \times \frac{10^n - 1}{9}$.

These representations are useful in various mathematical problems involving repeating digit numbers.