The sum of n terms of the series 4+44+444+.... is ?
Answers
Answered by
183
4+44+444+....... upto n terms.
4(1+11+111+........upto n terms)
4/9(9+99+999+.........upto n terms)
4/9[(10-1)+(10²-1)+...... upto n terms]
4/9[10+10²+.........+10ⁿ}-(1+1+1+....n times)]
(4/9) { 10 [ ( 10ⁿ - 1 ) / ( 10 - 1 ) ] - n(1) }
(4/9) [ (10/9) ( 10ⁿ - 1 ) - n ]...................
4(1+11+111+........upto n terms)
4/9(9+99+999+.........upto n terms)
4/9[(10-1)+(10²-1)+...... upto n terms]
4/9[10+10²+.........+10ⁿ}-(1+1+1+....n times)]
(4/9) { 10 [ ( 10ⁿ - 1 ) / ( 10 - 1 ) ] - n(1) }
(4/9) [ (10/9) ( 10ⁿ - 1 ) - n ]...................
Ekman:
try to solve it in ur notebook, if problem still exists, then i will ready to help
see, in: 4/9[10+10²+.........+10ⁿ}-(1+1+1+....n times)]
(1+1+1+....n times) gives n*1 got it?
in : (4/9) { 10 [ ( 10ⁿ - 1 ) / ( 10 - 1 ) ] - n(1) } step
in: 4/9[10+10²+.........+10ⁿ}-(1+1+1+....n times)] step
[10+10²+.........+10ⁿ} forms gp
and sum of gp is: rⁿ-1/r-1, where r is first term of gp
thank u very much asking help
any other query, please do ask.....
thanx 4 ur help
Answered by
50
Answer:
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) + ... + (10ⁿ - 1) ]
= (4/9) [ ( 10 + 10² + 10³ + ... + 10ⁿ ) - ( 1 + 1 + 1 + ... n times ) ]
= (4/9) { 10 [ ( 10ⁿ - 1 ) / ( 10 - 1 ) ] - n(1) }
= (4/9) [ (10/9) ( 10ⁿ - 1 ) - n ] ................................. Ans-
Similar questions