Question

if 1^2+2^2+.......+9^2=285 then the value of( 0.11)^2+(0.22)^2+.....(0.99)^2 is

Answer 1

Answer is = 3.4485

Hope it helps you

Answer 2

Solution:

→$1^{2} + 2^{2} +........+9^{2} = 285$

→Now, $=( 0.11)^2+(0.22)^2+.....(0.99)^2= [0.11 \times 1]^2 + [0.11 \times2]^2 +[0.11 \times 3]^3+........+ [0.11 \times 9]^2\\\\= (0.11)^2[  1^2+2^2+.......+9^2]\\\\= 0.0121 \times 285$

→∵[1^2+2^2+.......+9^2=285]

→ 3.4485 (Answer)