Question

The sum of n terms of the series 4 + 44 + 444 + ......... is

• A.
  (4/81) [${10}^{n+1}-9n-1]$
• B.
  (4/81) [${10}^{n-1}-9n-1]$
• C.
  (4/81) [${10}^{n+1}-9n-10]$
• D.
  (4/81) [${10}^n-9n-10]$

Answer 1

The correct option is C

(4/81) [${10}^{n+1}-9n-10]$
4 + 44 + 444 .........

If we see first term of series = 4

Hence, sum upto first term is also = 4

Put, n=1 (first term) only option (c) satisfies

$=\frac{4}{81}[{10}^{n+1}-9n-10]$

$=\frac{4}{81}[{10}^{1+1}-9\times 1-10]$

$=\frac{4}{81}\times 81=4$

Hence, option (c) is correct answer. Always solve these questions by putting values.

Alternative Solution

4+44+444.............

$\frac{4}{9}[9+99+\mathrm{999......}]=\frac{4}{9}\left[\right(10-1)+({10}^2-1)+({10}^3-1).....]=\frac{4}{9}\left[\right(10+{10}^2+{10}^3..{.10}^n)-(1+1+\mathrm{1...}\left)\right]=\frac{4}{81}[{10}^{n+1}-9n-10]$