NDA2021MathematicsSequences and SeriesActual
If p=(1111 up to n digits ) , then what is the value of 9 p^2+p ?
Options
- A10^n p
- B2 p 10^n
- C10n p-1
- D10^n p+1
Correct answer
A. 10^n p
Step-by-step solution
aligned & P=[1111 up to n digits ] & P= 1 9 [9999 upto n digits ] & = 1 9 [10^n-1 ] & 9 P ^2+ P = 1 9 [(10 n)^2-2 10^n+1 ]+ 1 9 [10^n-1 ] & = 1 9 [ (10^n )^2-10^n ]=10^n 1 9 [10^n-1 ] & =10^n P aligned