Binomial Theorem and Its Simple Applications

1Binomial expansion for positive integral n

For a positive integer n, (a+b)^n=\sum_{r=0}^{n}{}^nC_r\,a^{n-r}b^r. The sum runs from r=0 to r=n, so there are n+1 terms, not n. The usual slip is counting the index as the term count. The coefficients {}^nC_r are symmetric about the middle: {}^nC_r={}^nC_{n-r}, which is why the Pascal row reads the same forwards and backwards.

Write the sum \sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^{r} term by term — the figure would only re-typeset that identity.

Reading the expansion

1. Write the sum(a+b)^n expands as \sum_{r=0}^{n}{}^nC_r\,a^{n-r}b^r — one term for each r.
2. Count termsr runs through n+1 values, so the expanded polynomial has n+1 terms.
3. Use symmetry{}^nC_r={}^nC_{n-r} pairs the ends of the row; the greatest coefficient sits in the middle.

2General term, and the term independent of x

The (r+1)-th term of (a+b)^n is T_{r+1}={}^nC_r\,a^{n-r}b^r. The term number is one ahead of the exponent index, because the exponent index starts at zero: T_1 is the a^n term. To locate a specific power of x, or the term independent of x, write T_{r+1}, collect the total exponent of x, set that exponent to the required value and solve for r.

Figure. Write T_{r+1}={}^9C_r\,x^{18-3r} for (x^2+1/x)^9. The independent term needs exponent 18-3r=0, so r=6, and {}^9C_6=84.

Finding a required term

1. Write T_{r+1}Substitute the two parts of the binomial into T_{r+1}={}^nC_r\,a^{n-r}b^r.
2. Collect the power of xSimplify every factor that carries x into a single power x^{\ldots}.
3. Solve for rSet the exponent to the required value (zero for the independent term) and evaluate {}^nC_r.

3Middle term(s)

The middle of the expansion depends on whether n is even or odd. If n is even there is one middle term, the \left(\dfrac{n}{2}+1\right)-th. If n is odd there are two middle terms, the \left(\dfrac{n+1}{2}\right)-th and the \left(\dfrac{n+3}{2}\right)-th. The trap is treating every expansion as if it had a single middle term.

Middle means an index in a finite list of n+1 terms — odd n gives one middle term, even n gives two.

4Sums of binomial coefficients

Substituting specific values of a and b turns the expansion into identities about the coefficients alone. Put a=b=1 and the left side is 2^n, so the sum of all coefficients is 2^n. Put a=1, b=-1 and the alternating sum is 0; combining that with the full sum isolates the even-position and odd-position sums, each equal to 2^{n-1}. The greatest coefficient in the row is the middle one.

Figure. Put a=b=1 to get every coefficient summing to 2^n; for (x+y)^8 that is 256. Put a=1, b=-1 to get alternating sum 0. Adding those identities splits the row into even-position and odd-position sums, each 2^{n-1}.

Isolating even and odd sums

1. Full sumSet a=b=1: (1+1)^n=2^n=\sum{}^nC_r.
2. Alternating sumSet a=1, b=-1: (1-1)^n=0=\sum(-1)^r{}^nC_r.
3. Add or subtractAdding the two identities doubles the even-position sum; subtracting doubles the odd-position sum. Each is 2^{n-1}.

5Greatest term in the expansion

"Greatest coefficient" and "greatest term" are different questions. The greatest coefficient is the middle {}^nC_r, settled by symmetry. The greatest term also depends on the ratio b/a: compare consecutive terms by \dfrac{T_{r+1}}{T_r}=\dfrac{n-r+1}{r}\cdot\dfrac{b}{a} and find where that ratio crosses 1. Terms grow while the ratio is greater than 1 and shrink once it falls below 1.

Figure. Greatest term is not the same as greatest coefficient. Form T_{r+1}/T_r=((n-r+1)/r)\cdot(b/a) and read the peak where that ratio crosses 1: terms grow while the ratio is above 1 and shrink once it falls below.

Where the terms peak

1. Form the ratioWrite \dfrac{T_{r+1}}{T_r}=\dfrac{n-r+1}{r}\cdot\dfrac{b}{a} from the general term.
2. Solve \dfrac{T_{r+1}}{T_r}\ge 1The inequality marks every r for which the next term is at least as large as the current one.
3. Read the peakThe greatest term (or two equal greatest terms) sits where the ratio crosses 1.

6Linear approximation (1+x)^n\approx 1+nx

When |x| is much smaller than 1, the binomial expansion of (1+x)^n is dominated by its first two terms: (1+x)^n\approx 1+nx. Higher powers x^2, x^3, \ldots are negligible beside x only while that increment stays tiny. The same expansion, written with a convenient multiple plus remainder — (\text{multiple}+r)^n — is the standard route to last digits and remainders on exam papers.

Figure. Near x=0, (1+x)^3 and its linearisation 1+3x share value and slope; the gap grows as |x| leaves the tiny-increment regime where higher powers stay negligible.