Question

The value of $(1.02{)}^4+(0.98{)}^4$ upto three places of decimal is

• A. $2.004$
• B. $2.003$
• C. $2.04$
• D. $2.0004$

Answer 1

The correct option is A $2.004$

${\left(1.02\right)}^4+{\left(0.98\right)}^4$

$={(1+0.02)}^4+{(1-0.02)}^4$

$=[{}^4{C}_0{\left(1\right)}^4+{}^4{C}_1{\left(1\right)}^{4-1}{\left(0.02\right)}^1+{}^4{C}_2{\left(1\right)}^{4-2}{\left(0.02\right)}^2+{}^4{C}_3{\left(1\right)}^{4-3}{\left(0.02\right)}^3+{}^4{C}_4{\left(1\right)}^{4-4}{\left(0.02\right)}^4]+[{}^4{C}_0{\left(1\right)}^4-{}^4{C}_1{\left(1\right)}^{4-1}{\left(0.02\right)}^1+{}^4{C}_2{\left(1\right)}^{4-2}{\left(0.02\right)}^2-{}^4{C}_3{\left(1\right)}^{4-3}{\left(0.02\right)}^3+{}^4{C}_4{\left(1\right)}^{4-4}{\left(0.02\right)}^4]$

$=2[{}^4{C}_0+{}^4{C}_2\left(0.0004\right)+{}^4{C}_4\left(0.00000016\right)]$

$=2\times (1+0.0024+0.00000016)$

$=2\times 1.002$

$=2.004$

Hence the value of given expression upto three places of decimal is $2.004$.