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Find the nth term of the following sequence:
2 + 22 + 222 + ……….. Tn
2 + 22 + 222 + ……….. Tn
- A2(10n – 1)
- B2n(10n – 1)
- C\(\frac{2}{9}\left( {{{10}^n} - 1} \right)\)
- D\({\left( {\frac{2}{9}} \right)^n}\left( {{{10}^n} - 1} \right)\)
Answer not set
Discussion 1
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CCommunity
Let us try to make sense of the data given,
A1 = 2 = 2 × 1
A2 = 22 = 2 × 11
A3 = 222 = 2 × 111
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An = 2222…..(n times) = 2 × 1111…..(n times)
Looking at the options given, the first and the second are straightaway ruled out, because they do not give us the required answer.
Looking at the fourth option, let us try n = 2,
$({\left( {\frac{2}{9}} \right)^n}\left( {{{10}^n} - 1} \right) = \frac{4}{{81}}\left( {100 - 1} \right) = \frac{{44}}{9})$
This is also not the right answer.
Now, look at option 3, and trying for n = 3,
$(\frac{2}{9}\left( {{{10}^n} - 1} \right) = \frac{2}{9}\left( {1000 - 1} \right) = 222)$
This is giving us the right result.
A1 = 2 = 2 × 1
A2 = 22 = 2 × 11
A3 = 222 = 2 × 111
.
.
.
.
An = 2222…..(n times) = 2 × 1111…..(n times)
Looking at the options given, the first and the second are straightaway ruled out, because they do not give us the required answer.
Looking at the fourth option, let us try n = 2,
$({\left( {\frac{2}{9}} \right)^n}\left( {{{10}^n} - 1} \right) = \frac{4}{{81}}\left( {100 - 1} \right) = \frac{{44}}{9})$
This is also not the right answer.
Now, look at option 3, and trying for n = 3,
$(\frac{2}{9}\left( {{{10}^n} - 1} \right) = \frac{2}{9}\left( {1000 - 1} \right) = 222)$
This is giving us the right result.