Binomial Theorem for Positive Integral Index

Building up the pattern. Direct multiplication gives: $(a+b)^0=1$; $(a+b)^1=1a+1b$; $(a+b)^2=1a^2+2ab+1b^2$; $(a+b)^3=1a^3+3a^2b+3ab^2+1b^3$; $(a+b)^4=1a^4+4a^3b+6a^2b^2+4ab^3+1b^4$. Arranging the coefficients of these five expansions in a triangular array gives Pascal's triangle, and each coefficient can be written as a combination $^nC_r$ (see the table sidecar for both forms laid out row by row). Recognising the coefficients as $^nC_r$ is what lets the pattern be stated once, for a general power $n$, instead of re-deriving it for each new exponent by repeated multiplication.

The theorem. If $a,b\in R$ and $n\in N$, then $(a+b)^n = {}^nC_0\,a^n b^0 + {}^nC_1\,a^{n-1}b^1 + {}^nC_2\,a^{n-2}b^2 + \cdots + {}^nC_n\,a^0b^n.$

Proof, by the Principle of Mathematical Induction (from Section 4.1). Let $P(n)$ be the statement above. Step (I). At $n=1$: L.H.S. $=a+b$; R.H.S. $={}^1C_0a^1b^0+{}^1C_1a^0b^1=a+b$. Equal, so $P(1)$ true. Step (II). Assume $P(k)$: $(a+b)^k={}^kC_0a^kb^0+{}^kC_1a^{k-1}b^1+\cdots+{}^kC_ka^0b^k$. Step (III). We must prove $(a+b)^{k+1}={}^{k+1}C_0a^{k+1}b^0+{}^{k+1}C_1a^kb^1+\cdots+{}^{k+1}C_{k+1}a^0b^{k+1}$. Write L.H.S. $=(a+b)(a+b)^k=(a+b)\left[{}^kC_0a^kb^0+{}^kC_1a^{k-1}b^1+\cdots+{}^kC_ka^0b^k\right]$ by Step II. Distributing the factor $(a+b)$ over every term and then collecting terms with the same power of $a$ and $b$, each interior coefficient becomes a sum $\left({}^kC_r+{}^kC_{r-1}\right)$ of two neighbouring coefficients from row $k$ of Pascal's triangle. Pascal's rule — a standard combinatorial identity — states $^kC_r+{}^kC_{r-1}={}^{k+1}C_r$, and the two end coefficients satisfy $^kC_0=1={}^{k+1}C_0$ and $^kC_k=1={}^{k+1}C_{k+1}$. Substituting these gives exactly $(a+b)^{k+1}={}^{k+1}C_0a^{k+1}b^0+{}^{k+1}C_1a^kb^1+\cdots+{}^{k+1}C_{k+1}a^0b^{k+1}=$ R.H.S., so $P(k+1)$ is true. Step (IV). By the Principle of Mathematical Induction, $P(n)$ is true for all $n\in N$.

Remarks. (1) The expansion of $(a+b)^n$ has $n+1$ terms. (2) The first term is $a^n$ and the last term is $b^n$. (3) In every term the exponents of $a$ and $b$ add up to $n$. (4) Moving from one term to the next, the exponent of $a$ decreases by $1$ while the exponent of $b$ increases by $1$. (5) Coefficients of terms equally far from the two ends of the expansion are equal — the coefficients are symmetric, matching the symmetry of Pascal's triangle. (6) For subtraction: $(a-b)^n={}^nC_0a^nb^0-{}^nC_1a^{n-1}b^1+{}^nC_2a^{n-2}b^2-\cdots+(-1)^n{}^nC_na^0b^n$ — the sign simply alternates term by term, starting positive.

Solved Example 1. Expand $(x^2+3y)^5$. Solution. Here $a=x^2,b=3y,n=5$. Using $^5C_0={}^5C_5=1$, $^5C_1={}^5C_4=5$, $^5C_2={}^5C_3=10$: $(x^2+3y)^5=1(x^{10})+5(x^8)(3y)+10(x^6)(9y^2)+10(x^4)(27y^3)+5(x^2)(81y^4)+1(243y^5)=x^{10}+15x^8y+90x^6y^2+270x^4y^3+405x^2y^4+243y^5$.

Solved Example 2. Expand $\left(2x-\dfrac{y}{2}\right)^5$. Solution. Here $a=2x,b=y/2,n=5$. $\left(2x-\dfrac{y}{2}\right)^5=32x^5-5(16x^4)\left(\dfrac{y}{2}\right)+10(8x^3)\left(\dfrac{y^2}{4}\right)-10(4x^2)\left(\dfrac{y^3}{8}\right)+5(2x)\left(\dfrac{y^4}{16}\right)-\dfrac{y^5}{32}=32x^5-40x^4y+20x^3y^2-5x^2y^3+\dfrac{5}{8}xy^4-\dfrac{y^5}{32}$.

Solved Example 3. Expand $(\sqrt5+\sqrt3)^4$. Solution. Here $a=\sqrt5,b=\sqrt3,n=4$, and $^4C_0={}^4C_4=1$, $^4C_1={}^4C_3=4$, $^4C_2=6$: $(\sqrt5+\sqrt3)^4=1(25)+4(5\sqrt5)(\sqrt3)+6(5)(3)+4(\sqrt5)(3\sqrt3)+1(9)=25+20\sqrt{15}+90+12\sqrt{15}+9=124+32\sqrt{15}$.

Solved Example 4. Evaluate $(\sqrt2+1)^5-(\sqrt2-1)^5$. Solution. Expand each with $n=5$ and the coefficients $1,5,10,10,5,1$: $(\sqrt2+1)^5=4\sqrt2+20+20\sqrt2+20+5\sqrt2+1$ and $(\sqrt2-1)^5=4\sqrt2-20+20\sqrt2-20+5\sqrt2-1$ (grouping the $\sqrt2$-terms and the plain-number terms separately). Subtracting cancels the plain-number terms and doubles the $\sqrt2$-terms' coefficients where they matched; the net result is $2(20+20+1)=82$.

Solved Example 5 (Activity). Using the binomial theorem, find the value of $(99)^4$. Solution. Write $99=100-1$, so $(99)^4=(100-1)^4$. Using $^4C_0={}^4C_4=1$, $^4C_1={}^4C_3=4$, $^4C_2=6$: $(100-1)^4=1(100)^4-4(100)^3+6(100)^2-4(100)+1(1)=100000000-4000000+60000-400+1=96059601$. …