Question

9,99, 999, 9999 ....... t100, Find the Sum ?​

Answer 1

9 + 99 + ... +t100

By taking 9 common

9(1 + 11 + 111 + ...... + t 100)

= 9 { (0 + 1) + (10 + 1) + (100 + 1) ....... + ( 10¹⁰ + 1) }

= 9 [ (0 + 10 + 100 + ...... + 10¹⁰) + (1 + 1 + ...... + 1) ]

= 9 [ 0 + 10 + ..... + 10¹⁰] + 9 [ 100(1)]

Let A = 0 + 10 + ..... + 10¹⁰

a = 10 , r = 10, n = 99

$</p><p>s \: = \frac{a(r {}^{n} - 1) }{r - 1}</p><p>$

A = 10 ( 10⁹⁹ - 1)/ 9

Now,

9 + 99 + ... +t100

= 9 [ 0 + 10 + ..... + 10¹⁰] + 9 [ 100(1)]

= A + 900

= 10 ( 10⁹⁹ - 1)/ 9 +900

Hope this helps!

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