Question

If​ ​ 13​ ​ + ​ ​ 23+​ ​ 33+​ ​ ..........+​ ​ 93​ ​ = ​ ​ 2025​ ​ ,then​ ​ the​ ​ value​ ​ of​ ​ (0.11)3​ ​ + ​ ​ (0.22)3​ ​ + ​ ​ ....+​ ​ (0.99)3​ ​ is​ ​ close​ ​ to:

Answer 1

so a=13
difference = d=23-13=10
a+(n-1)*10= 13+10n-10=3+10n=2025
or 10n=2022
or n = 202.2
frm here u got n
now in the next series a=.11
difference =.11 ...try this method to get the solution. ..

Answer 2

The value of (0.11)3+(0.22)3+(0.33)3+.......+(0.99)3 = 20543.52 .

•Given series is 13+23+33+.....+99 = 2025.

• from the above series ,it is in the form of a,a+d,a+2d,....

•Therefore,a = 13,

a+d = 23

• then,d= 10

•We know the formula for Sn= a+(n-1)d

•Where Sn=2025.

2025 = 13+(n-1)10

2025 = 13+10n-10

2025 = 3+10n

10n = 2022

n = 202.2 ...(1)

• the given second series is (0.11)3+(0.22)3+(0.33)3+....(0.99)3

• it is also in the form of a,a+d,a+2d...

• a = 0.33 ,a+d = 0.66

a+d = 0.66

0.33+d = 0.66

d = 0.33

•To find the sum of series Sn,we have n=202.2

•Therefore,sum of n numbers is n(n+1)/2

(202.2)(202.2+1)/2

=(101.1)(203.2) = 20543.52

•Therefore,the sum of (0.11)3+(0.22)3+(0.33)3+...+(0.99)3 = 20543.52.