# Find 2 + 22 + 222 + 2222 + … upto n terms. - Mathematics and Statistics

Sum

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

#### 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 + 100 + ... upto n terms) – (1 + 1 + 1 ... n terms)]

Since, 10, 100,1000, … n terms are in G.P. with a= 10, r = 100/100 = 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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#### APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 11th Standard Maharashtra State Board
Chapter 4 Sequences and Series
Miscellaneous Exercise 4 | Q 8 | Page 64