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

Answers (1)

4 + 44 + 444 + ...n terms

= 4 [ 1 + 11 + 111 + ... n terms ]

= (4/9) [ 9 + 99 + 999 + ...  n terms ]

= (4/9) [ (10 - 1) + (102 - 1) + (103 - 1) + ... + (10n - 1) ]

= (4/9) [ ( 10 + 102 + 10n + ... + 10n ) - ( 1 + 1 + 1 + ... n times ) ]

= (4/9) { 10 [ ( 10n - 1 ) / ( 10 - 1 ) ] - n(1) }

= (4/9) [ (10/9) ( 10n - 1 ) - n ] 

Posted by

Ravindra Pindel

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