Respuesta :
Answer:
Sum of the digits of the number
[tex]= ((\frac{60((10)^{100}-1) }{81}) - \frac{600}{9} )[/tex]
Step-by-step explanation:
Step(i):-
Given series
6+66+666+6666 + ... +666...66 up to 100 digits
Taking common '6'
6 ( 1 + 11 +111+ 1111+1111+.................11111....11 100 digits)
Multiply '9' and divisible by'9'
[tex]\frac{6}{9} X9( 1+11 +111 +1111 +1111+ ..........11111..up to .100 digits )[/tex]
Multiply inside '9'
[tex]\frac{6}{9}( 9+99 +999 +9999 +.....+ ..........9999..up to .100 digits )[/tex]
[tex]\frac{6}{9}( (10-1)+(100-1)+(1000-1) +(10,000-1)+.....+ ............up to .100 digits )[/tex]
[tex]\frac{6}{9}( (10)+(100)+(1000) +(10,000)+.....+ ............up to .100 digits ) - ( 1+1+1+1+........up to 100 digits)[/tex]
Step(ii):-
we know that sum of geometric series
[tex]S_{n} = \frac{a(r^{n}-1) }{r-1}[/tex]
we know that
a + a + a+........n terms = n a
[tex]\frac{6}{9}( (10)+(10)^{2} +(10)^{3} +(10)^4+.....+ ............up to .100 digits - (100(1))[/tex] ....(i)
The sum of the 100 digits in geometric series
[tex]S_{100} = \frac{10((10)^{100}-1) }{10-1}[/tex]
Now the equation (i)
The sum of the digits of the number
[tex]= \frac{6}{9} ((\frac{10((10)^{100}-1) }{10-1}) - 100)[/tex]
Final answer :-
Sum of the digits of the number
[tex]= ((\frac{60((10)^{100}-1) }{81}) - \frac{600}{9} )[/tex]
