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NDA 2021
Exam
NDA
Year
2021
Subject
Mathematics
Chapter
Sequences and Series

NDA 2021 · Mathematics · Sequences and Series

If p=(1111 up to n digits ) , then what is the value of 9 p^2+p ?

If p=(1111…p=(1111 \ldots up to nn digits )), then what is the value of 9p2+p9 p^2+p ?

Options

  1. A10np10^n p
  2. B2p⋅10n2 p \cdot 10^n
  3. C10np−110n p-1
  4. D10np+110^n p+1

Correct answer

A. 10np10^n p

Step-by-step solution

P=[1111… up to  n digits ]P=19[9999… upto n digits ]=19[10n−1]∴9P2+P=19[(10n)2−2⋅10n+1]+19[10n−1]=19[(10n)2−10n]=10n⋅19[10n−1]=10n⋅P\begin{aligned} & P=[1111 \ldots \text{ up\text{ to } } n \text{ digits }] \\ & P=\frac{1}{9}[9999 \ldots \text{ upto } n \text{ digits }] \\ & =\frac{1}{9}\left[10^n-1\right] \\ & \therefore 9 \mathrm{P}^2+\mathrm{P}=\frac{1}{9}\left[(10 n)^2-2 \cdot 10^n+1\right]+\frac{1}{9}\left[10^n-1\right] \\ & =\frac{1}{9}\left[\left(10^n\right)^2-10^n\right]=10^n \cdot \frac{1}{9}\left[10^n-1\right] \\ & =10^n \cdot \mathrm{P}\end{aligned}
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