Q.Use the binomial theorem to find the approximate value of $(1.02)^{5}$, correct to three decimal places, keeping terms only up to the $y^2$ term.

Writing $1.02=1+0.02$ turns this into a binomial expansion where the terms shrink rapidly, so keeping just the first three terms already gives a good decimal approximation.

Write $1.02 = 1+0.02$, so $(1.02)^5 = (1+y)^5$ with $y=0.02$. By the binomial theorem, $(1+y)^5 = 1+5y+10y^2+10y^3+5y^4+y^5$.

Since $y=0.02$ is small, the terms shrink fast: $y^2=0.0004$, $y^3=0.000008$, and beyond that the terms are negligible for a 3-decimal-place answer. Keeping terms only up to $y^2$, as asked:

$(1.02)^5 \approx 1+5(0.02)+10(0.02)^2 = 1+0.1+10(0.0004) = 1+0.1+0.004 = 1.104$ …

Forgetting the coefficient $^{5}C_{2}=10$ on the $y^2$ term (writing just $y^2$ instead of $10y^2$), which would understate the approximation; also stopping at the $y^1$ ter …