5+55+555+5555+.......nth term sum=?
Answers
Answered by
171
first of all take 5 common like this
5(1+11+111+....n)
then multiply and divide by 9
5/9*9(1+11+111+....n)
5/9(9+99+999.....n)
now we can write 9 like this
5/9[(10-1)+(10*10-1)+(10*10*10-1).....n)
now
5/9(10+10*10+10*10*10....n)-[1+1+1+1....n)
so here a=10, r=10, r>1
so use this formulae
S=a(r to the power n - 1)/r-1
= 5/9[10(10 to the power n - 1)/10-1] - n
= 5/9[10/9(10 to the power n - 1) - n
= 50/81(10 to the power n -1) - 5/9n this is the answer have fun
5(1+11+111+....n)
then multiply and divide by 9
5/9*9(1+11+111+....n)
5/9(9+99+999.....n)
now we can write 9 like this
5/9[(10-1)+(10*10-1)+(10*10*10-1).....n)
now
5/9(10+10*10+10*10*10....n)-[1+1+1+1....n)
so here a=10, r=10, r>1
so use this formulae
S=a(r to the power n - 1)/r-1
= 5/9[10(10 to the power n - 1)/10-1] - n
= 5/9[10/9(10 to the power n - 1) - n
= 50/81(10 to the power n -1) - 5/9n this is the answer have fun
Answered by
45
Answer:
This is not a GP
Sn = 5+55+555+5555+........ to n terms
= 5/9(9+99+999+...... to n terms)
= 5/9[(10-1) + (10^2-1) +(10^3-1) + ...... to n terms]
=5/9[(10+10^2 + 10^3 + ...... n terms)-(1+1+1+... n terms)]
=5/9[10(10^n--1)/10-1 -- n]
= 5/9[10(10^n--1)/9 --- n]
Step-by-step explanation:
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