If p = (1111 up to n digits), then what is the value of 9p² + p?

Correct Answer is : 10np10ⁿ p10np

Correct Answer is : 10np10ⁿ p10np

Test Index

UPSC NDA 2 2021 Mathematics Solved Paper

Question : 5 of 120

Marks: +1, -0

p = (1111\ldots

9p^{2} + p

Solution:

Concept:

p

is a repunit number (all ones) which can be expressed as the sum of a geometric series.

Explanation:

p = 1111\ldots

(n digits) can be written as:

p = 1 + 10 + 10^{2} + 10^{3} + \dots + 10^{n-1}

This is a geometric progression with first term

a = 1

, common ratio

r = 10

, and

n

terms.

Using the sum formula

S_n = \frac{a(r^{n} - 1)}{r - 1}

, we get:

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

Now evaluate

9p^{2} + p

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

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

= \frac{(10^{n} - 1)(10^{n} - 1 + 1)}{9}

= \frac{(10^{n} - 1) \cdot 10^{n}}{9}

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

Answer:

9p^{2} + p = 10^{n}p

, which corresponds to option A.

Go to Question:

Prev Question

Next Question