Q.Expand $(x+1)^6$ using the binomial theorem.

The Binomial Theorem for a positive integral index gives a single formula for expanding $(a+b)^n$ for any natural number $n$, without multiplying the bracket out by hand $n$ times: $(a+b)^n = {}^nC_0\,a^nb^0 + {}^nC_1\,a^{n-1}b^1 + {}^nC_2\,a^{n-2}b^2 + \cdots + {}^nC_n\,a^0b^n,$ where the coefficients $^nC_0,{}^nC_1,\ldots,{}^nC_n$ are exactly the numbers that appear in Pascal's triangle. The theorem itself is proved using the Principle of Mathematical Induction, with Pascal's rule ($^kC_r+{}^kC_{r-1}={}^{k+1}C_r$) doing the work of the inductive step. A few structural facts follow immediately from the formula and are worth remembering as a checklist: the expansion always has exactly $n+1$ terms; the first term is $a^n$ and the last is $b^n$; in every term the exponents of $a$ and $b$ add up to $n$; moving along the expansion, the exponent of $a$ steadily falls by one while that of $b$ steadily rises by one; and the coefficients read the same from either end (they are symmetric), which mirrors the left-right symmetry of Pascal's triangle. For the subtraction case, $(a-b)^n$ has the same coefficients but with alternating signs, starting positive. This theorem underlies a large family of exam techniques: expanding an explicit power like $(x^2+3y)^5$; evaluating expressions like $(\sqrt3+\sqrt2)^4$ where the surds partly cancel between a $+$ and a $-$ version; approximating a number like $(2.02)^5$ or $(9.9)^3$ by writing it as $(a+b)^n$ with a small $b$ and truncating the series after a few decimal-significant terms; and instantly evaluating an expression that is secretly a full binomial expansion in disguise, such as $(2x-1)^4+4(2x-1)^3(3-2x)+\cdots+(3-2x)^4$, which collapses to $[(2x-1)+(3-2x)]^4$.