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Answer the following: Find 2 + 22 + 222 + 2222 + ... upto n terms - Mathematics and Statistics

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Question

Answer the following:

Find 2 + 22 + 222 + 2222 + ... upto n terms

Sum

Solution

Sn = 2 + 22 + 222 +… upto n terms

= 2(1 + 11 + 111 + … upto n terms)

= `2/9` (9 + 99 + 999 + ... upto n terms)

= `2/9` [(10 − 1) + (100 − 1) + (1000 − 1) + ... upto n terms]

= `2/9` [(10 + 100 + 1000 + ... upto n terms) − (1 + 1 + 1 + ... n times)]

But 10, 100, 1000, … n terms are in G.P.

with a = 10, r = `100/10` = 10

∴ Sn =  `2/9[10((10^"n" - 1)/(10 - 1)) - "n"]`

= `2/9[10/9(10^"n" - 1) - "n"]`

∴ Sn = `2/81[10(10^"n" - 1) - 9"n"]`

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Arithmetico Geometric Series
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Chapter 2: Sequences and Series - Miscellaneous Exercise 2.2 [Page 41]

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Balbharati Mathematics and Statistics 2 (Arts and Science) [English] 11 Standard Maharashtra State Board
Chapter 2 Sequences and Series
Miscellaneous Exercise 2.2 | Q II. (8) | Page 41

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