Question

1^2 +2^2+3^2+......+9^2 =285 to (0.11)^2+(0.22)^
2+......(0.99)^2 =?

Answer 1

Answer

(0.11)^2+(0.22)^2+......(0.99)^2  =3.4485

Explanation

It is given that,

1^2 +2^2+3^2+......+9^2 =285

To find  (0.11)^2+(0.22)^2+......(0.99)^2

(0.11)^2+(0.22)^2+......(0.99)^2  can be written as,

(0.11 x 1)^2+(0.11 x 2)^2+......(0.11 x 9)^2

 =(0.11^2 x 1^2)+(0.11^2 x 2^2)+......(0.11^2 x 9^2)

(0.11^2 )( 1^2 +2^2+3^2+......+9^2 )   because  0.11 is common in all terms

We know that 1^2 +2^2+3^2+......+9^2  = 285

(0.11^2 )( 1^2 +2^2+3^2+......+9^2 ) = (0.11^2 ) x 285 = 0.0121 x 285 = 3.4485

Answer 2

Given 1^2 + 2^2 + 3^2...................+9^2 = 285

We have to find the value of

 0.11^2 + 0.22^2 + ...................+0.99^2

We can write this as

 (0.11 x 1)^2 + (0.11 x 2)^2.............+(011 x 9)^2

 0.11^2 x 1^2 + 0.11 x 2^2............+0.11^2 x 9^2

By taking 0.11^2 as common factor

0.11^2(1^2 + 2^2 .................+9^2)

0.11^2 x 285 (given 1^2 + 2^2..........+9^2 = 285)

= 0.0121 x 285 = 3.4485