Q.Using the binomial theorem, evaluate $(1.1)^5$ correct to $3$ decimal places.

Beyond pure expansion, the binomial theorem gives three standard problem-solving tools. First, approximation: when $x$ is small, $(1+x)^n=\sum{}^{n}C_r x^r$ can be truncated after the first few terms to compute a power like $(1.02)^6$ to several decimal places without a calculator. Second, comparison: since every term of $(1+x)^n$ is positive for $x>0$, the two-term truncation $(1+x)^n\ge1+nx$ is a valid lower bound, often enough by itself to decide which of two quantities is larger. Third, divisibility: writing $a=1+b$ and expanding $(1+b)^n$ shows that $a^n-nb-1=\sum_{r=2}^{n}{}^{n}C_r b^r$, in which every term carries a factor of $b^2$, giving a clean proof that expressions such as $9^n-8n-1$ (divisible by $64$) or $11^n-10n-1$ (divisible by $100$) hold for every positive integer $n$. Setting $a=b=1$ or $a=1,b=-1$ in the theorem also gives the two standard identities $\sum{}^{n}C_r=2^n$ and the alternating sum $=0$.

These applications come from the NCERT Class 11 Mathematics chapter on Binomial Theorem, a chapter tested in CBSE boards and a genuine favourite in JEE Main for its approximation and divisibility-proof questions. It is exactly what a search for "binomial theorem applications class 11 maths" or "binomial theorem divisibility proof JEE" is looking for.