[tex]\mathfrak{\huge{\red{\underline{{Answer:-}}}}}[/tex]

To add the given expressions, we need to expand them first:

x(x+1)₂ = x² + x

x(x+1)₂ = x² + xx(x+2) = x² + 2x

x(x+1)₂ = x² + xx(x+2) = x² + 2xx(x+3) = x² + 3x

Now we can write the sum as follows:

x(x+1)₂ + x(x+2) + x(x+3) = x² + x + x² + 2x + x² + 3x

= 3x² + 6x

To evaluate the sum for x=1/2, we substitute x=1/2:

3x² + 6x = 3(1/2)² + 6(1/2) = 3/4 + 3 = 15/4

Therefore, the sum is 15/4 when x=1/2.

To evaluate the sum for x=0.2, we substitute x=0.2:

3x² + 6x = 3(0.2)² + 6(0.2) = 0.12

Therefore, the sum is 0.12 when x=0.2.

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[tex]\small\blue{Answer \: by \: ☞}\green{ \: Arya \: Aditya ✨}[/tex]