the sum of n terms of series 4 44 444 4444 44444 is 1 4 81 10 n 1 -9n-1

putting 1 we will get the 1st term that is 4,so only eqn 2 satisfies thus answer is(3) only
11 years agoHelpfull: Yes(4) No(3)
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) + (10^2 - 1) + (10^3 - 1) + ... + (10^n - 1) ]

= (4/9) [ ( 10 + 10^2 + 10^3 + ... + 10^n ) - ( 1 + 1 + 1 + ... n times ) ]

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

= (4/9) [ (10/9) ( 10^n - 1 ) - n ]
= (4/81)[10^(n+1)-9n-10]
11 years agoHelpfull: Yes(3) No(3)
4+44+444+4444+...=4(1+11+111+1111+ . . .)
Now using geometric series we can evaluate,
1+11+111+1111+...=1+(1*10+1)+{(1*10)*10+1}+ . .
If you replace the 10 with r, it reduces to a geometric series,
1/(r-1)^2 (r^(n+1)-(1+n)r +n)
Hence the result is 1/81{10^(n+1)-10(1+n) +n } =1/81{10^(n+1)-10-10 n+n}
=1/81{10^(n+1)-9 n-10}. . According to our requirement in the question,multiply by 4 . . So the answer is (3) :) ;)
11 years agoHelpfull: Yes(1) No(2)
the correct ans. is 3 i.e. (4/81)[10^(n+1)-9n-10]
cause 4(1+11+111+1111+.............)
4/9[(1+11+111+......)*9]
4/9[(10+10^2+10^3+.....)-n]
it's in g.p. now aaplying formula of sum of n terms of gp we get the correct ans. as 3rd option
11 years agoHelpfull: Yes(0) No(3)

the sum of n terms of series 4+44+444+4444+44444+ ......is
(1) (4/81)[10^(n+1)-9n-1]
(2) (4/81)[10^(n-1)-9n-1]
(3) (4/81)[10^(n+1)-9n-10]
(4) (4/81)[10^n-9n-10]