Binomial theorem for positive integral indices

For any real numbers $a, b$ and any positive integer $n$, the binomial theorem states that $(a+b)^n$ expands as a sum of $n+1$ terms, with the power of $a$ falling from $n$ to $0$ and the power of $b$ rising from $0$ to $n$ across the sum:

where: $n$ is the (positive integer) index, $k$ runs from $0$ to $n$ over each term, and ${}^nC_k = \dfrac{n!}{k!(n-k)!}$ (also written $\binom{n}{k}$) is the binomial coefficient, the count of ways to choose $k$ objects from $n$. Three structural facts turn this abstract sum into something you can read off by eye. First, term count: $(a+b)^n$ always has $n+1$ terms, not $n$ — a favourite place to lose a mark. Second, symmetry: ${}^nC_k = {}^nC_{n-k}$, so the coefficients read the same forwards and backwards, e.g. row $4$ is $1, 4, 6, 4, 1$. Third, the sign flip: replacing $b$ by $-b$ gives $(a-b)^n = \sum_{k=0}^n (-1)^k\,{}^nC_k\,a^{n-k}b^k$, so the signs alternate $+,-,+,-,\dots$ while the coefficients themselves stay unchanged. Those coefficients are exactly the rows of Pascal's triangle: row $n$ lists ${}^nC_0, {}^nC_1, \dots, {}^nC_n$, and each entry equals the sum of the two entries diagonally above it in the row before, which is just the algebraic identity ${}^nC_{r-1} + {}^nC_r = {}^{n+1}C_r$. This makes the triangle a fast, error-free way to read off coefficients for small $n$ without ever computing a factorial: start from row $0$ (just the single entry $1$), and build each new row by adding adjacent pairs from the row above, with a $1$ fixed at each end. There is also a counting meaning worth keeping: thinking of $(a+b)^n$ as $n$ brackets $(a+b)(a+b)\cdots(a+b)$ multiplied out, each final term comes from choosing either $a$ or $b$ from every bracket, and ${}^nC_k$ counts how many of those choices put $b$ in exactly $k$ of the brackets — which is also why the coefficients must sum to $2^n$ when $a=b=1$. Historically, expansions for small $n$ were known in China, Persia and India centuries before Blaise Pascal, and the general statement for positive integral indices was fully established in Europe by the seventeenth century; what the theorem buys a student today is the same shortcut it always was — one formula instead of repeated multiplication.

Use the binomial theorem to evaluate $(101)^3$ without direct multiplication, then extend the same trick to $(99)^4$.

Step 1. Write $101 = 100+1$, so $(101)^3 = (100+1)^3$ with $a=100$, $b=1$, $n=3$. Row $3$ of Pascal's triangle gives coefficients $1, 3, 3, 1$.

Step 2. Expand: $(100+1)^3 = {}^3C_0(100)^3 + {}^3C_1(100)^2(1) + {}^3C_2(100)(1)^2 + {}^3C_3(1)^3 = 1{,}000{,}000 + 30{,}000 + 300 + 1$.

Step 3. Add the terms: $1{,}000{,}000 + 30{,}000 + 300 + 1 = 1{,}030{,}301$. So $(101)^3 = 1{,}030{,}301$.

Competitive stretch

Step 1. For $(99)^4$, write $99 = 100 - 1$, so $(99)^4 = (100-1)^4$ with $a=100$, $b=1$, $n=4$, using the alternating form since $b$ is being subtracted: row $4$ gives coefficients $1,4,6,4,1$ with signs $+,-,+,-,+$.

Step 2. Expand: $(100-1)^4 = (100)^4 - 4(100)^3(1) + 6(100)^2(1)^2 - 4(100)(1)^3 + (1)^4 = 100{,}000{,}000 - 4{,}000{,}000 + 60{,}000 - 400 + 1$.

Step 3. Add with signs: $100{,}000{,}000 - 4{,}000{,}000 + 60{,}000 - 400 + 1 = 96{,}059{,}601$, so $(99)^4 = 96{,}059{,}601$ — the same round-number trick, only with the alternating-sign form.

• Forgetting the sign pattern in $(a-b)^n$. All-positive signs is the default mistake; signs must alternate starting with $+$ for the $a^n$ term.
• Miscounting the number of terms. $(a+b)^n$ has $n+1$ terms, not $n$ — a common slip is stopping one term early.
• Swapping which power decreases. The power of $a$ falls from $n$ to $0$ while $b$'s power rises from $0$ to $n$; reversing this gives a wrong expansion even with correct coefficients.
• Sloppy coefficient arithmetic. Forgetting ${}^nC_0 = {}^nC_n = 1$, or miscalculating $\frac{n!}{k!(n-k)!}$ by hand instead of reading Pascal's triangle, is a frequent source of errors.

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