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

(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  we get

(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 )     ( since 0.11 is common in all terms)

it is given that 1^2 +2^2+3^2+......+9^2  = 285

Therefore,

(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

To find:

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

Solution:

Taking 0.11 as the common factor,

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

Multiplying ^2,

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

Grouping,

( 0.11^2 ) ( 1^2 + 2^2 + 3^2 + ... + 9^2 )  

Substituting,

(0.11^2 ) x 285

0.0121 x 285

3.4485

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