Question

555555...n terms. Change this term into G.P. then find the sum of this series?​

Answer 1

5+55+555+..........n

=5 (1+11+111+........n terms)

= 5 /9 (9+99+999+....... nterms )

= 5/9 ( (10 -1 ) + (100-1) + (1000-1) +.....nterms )

= 5/9 [(10 +100 + 1000 +......n terms (GP series ) ) - (1+1+1+1.......n terms )]

= 5/9 [ 10 ((10)n -1) / 10-1 ] - n ]

= 5/9 [ 10 (10n -1) /9 - n ]

= 5 /81 [ 10 (10n -1) - 9n ]

Answer 2

Answer:

Step-by-step explanation:

5 (1+11+111+........n terms)

= 5 /9 (9+99+999+....... nterms )

= 5/9 ( (10 -1 ) + (100-1) + (1000-1) +.....nterms )

= 5/9 [(10 +100 + 1000 +......n terms (GP series ) ) - (1+1+1+1.......n terms )]

= 5/9 [ 10 ((10)n -1) / 10-1 ] - n ]

= 5/9 [ 10 (10n -1) /9 - n ]

= 5 /81 [ 10 (10n -1) - 9n ]

hope it helps..!!