IF 1+10+100 +...UP TO 'n'TERMS IS EQUAL TO 10raised to the power of n-1 the whole divided by 9, then the sum of the series 4+44+444...UPTO 'n' terms is

Asked by  | 14th Apr, 2013, 12:20: PM

Expert Answer:


Please note, that in the question, it should be that 1+ 10 + 10^2 + 10^3 +...... n tems = [(10^n)-1]/9, as only then the LHS = RHS. 
 
 
4 + 44 + 444 + ... to n terms

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

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

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

= (4/9) [ ( 10 + 10² + 10³ + ... + 10? ) - ( 1 + 1 + 1 + ... n times ) ]
 
Since, sum of a GP = a(r^n -1)/(r-1) where a is the first term, r is the common ratio and n is the number of terms. 

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

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

Answered by  | 14th Apr, 2013, 05:40: PM

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