Binomial Theorem — NDA Mathematics Formulas and Traps

Coefficients & Specific Terms in the Expansion

The Binomial Theorem & the General Term

General term

T_{r+1} = \binom{n}{r}\, a^{\,n-r} b^{\,r}

Binomial Coefficients — C(n, r)

Binomial coefficient

\binom{n}{r} = \dfrac{n!}{r!\,(n-r)!}

Finding a Specific Term or Coefficient

Set the exponent, solve for r

\text{exponent of } x \text{ in } T_{r+1} = k \ \Rightarrow\ r \ \Rightarrow\ \text{coefficient}

The Middle Term

Middle term, n even

T_{\frac{n}{2}+1} = \binom{n}{n/2}\, a^{\,n/2} b^{\,n/2}

The Term Independent of x (Constant Term)

Constant term condition

\text{exponent of } x \text{ in } T_{r+1} = 0 \ \Rightarrow\ r

Conditions Linking Coefficients

First-three-terms shape

\binom{n}{1}a = (\text{2nd}),\quad \binom{n}{2}a^2 = (\text{3rd}) \ \Rightarrow\ \text{divide, solve } n

Counting Terms in Products and Powers

Distinct terms of a trinomial power

(a+b+c)^n \ \longrightarrow\ \binom{n+2}{2}\ \text{distinct terms}

Rational Terms & the General-Index Series

Rational-term test

\tfrac{n-r}{j} \in \mathbb{Z}\ \text{ and }\ \tfrac{r}{k} \in \mathbb{Z}

Common traps

Term number is r + 1, not r

T_{r+1}

uses

r

, but the term's POSITION is

r+1

. The 4th term has

r = 3

, not

r = 4

. Off-by-one here is the single most common binomial error.

Equal coefficients gives TWO cases

\binom{n}{a}=\binom{n}{b}

means

a=b

OR

a+b=n

. Students stop at

a=b

and miss the

a+b=n

solution (which is usually the one the question wants).

Collect every power of x first

A term like

\left(\tfrac{1}{x}\right)^r = x^{-r}

contributes a NEGATIVE power. Combine all the

x

-exponents into one expression before equating — forgetting a fractional or negative exponent is where most slips happen.

Odd n has two middle terms

For odd

n

the question may ask for "the middle term" expecting BOTH, or the ratio of the two. Count

n+1

terms and find the two central positions; don't report just one.

Multiply the bases before raising the power

(3x-y)^4(x+3y)^4 \ne

"add the term-counts". Combine the equal exponents into one base first —

[(3x-y)(x+3y)]^4

— then count.

BOTH exponents must be integers, not just one

A term is rational only when every surd disappears. Requiring only one of the two fractional exponents to be an integer over-counts — intersect the two conditions.

Sums of Binomial Coefficients

Learn this subtopic in the notes

Sum of All Coefficients — Put x = 1

Sum of coefficients = f(1)

\sum_{r=0}^{n}\binom{n}{r} = 2^n, \qquad \sum_{r=0}^{n}\binom{n}{r}c^{r} = (1+c)^n

Alternating & Odd/Even-Index Sums — Put x = −1

Alternating sum and the split

\sum_r (-1)^r \binom{n}{r} = 0, \qquad \text{even-sum} = \text{odd-sum} = 2^{\,n-1}

Weighted Sums via Differentiation

Index-weighted sum

\sum_{r=1}^{n} r\binom{n}{r} = n\,2^{\,n-1}

Pascal's Rule & Coefficient Identities

Pascal's rule (applied twice)

\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2} = \binom{n+2}{r}

Common traps

Sum of coefficients uses x = 1, not x = 0

f(0)

gives only the CONSTANT term;

f(1)

gives the sum of ALL coefficients. For a multivariable form set every variable to 1.

Differentiate first, substitute second

The factor of

r

only appears AFTER differentiating. Substituting

x=1

into

(1+x)^n

directly gives

2^n

, not the weighted sum — you must differentiate while

x

is still a variable.

Pascal's rule needs adjacent lower indices on the SAME n

\binom{n}{r} + \binom{n}{r-1}

(same top, consecutive bottom) combines to

\binom{n+1}{r}

.

\binom{n}{r} + \binom{n+1}{r}

(different tops) does NOT — don't force the rule on a mismatched pair.

Integer & Fractional Parts of Binomial Expressions

Learn this subtopic in the notes

The Conjugate Trick — Irrational Parts Cancel

Conjugate sum & product

(a+\sqrt{b})^n + (a-\sqrt{b})^n \in \mathbb{Z}, \qquad (a+\sqrt{b})^n(a-\sqrt{b})^n = (a^2-b)^n

Integer Part + Fractional Part

Fractional parts add to 1

f + f' = 1, \qquad f' = (a-\sqrt{b})^n

Common traps

Add the conjugate — don't expand the whole thing

Trying to expand

(a+\sqrt b)^{20}

term by term is hopeless. The intended move is always to bring in

(a-\sqrt b)^n

: its sum is an integer and its product is

(a^2-b)^n

.

The conjugate power IS the missing fractional part

f' = (a-\sqrt b)^n

is not just "small" — it is exactly

1 - f

. Treating

f'

as negligible or zero loses the relation

f + f' = 1

the question is built on.

Remainders & Divisibility via Binomial Expansion

Learn this subtopic in the notes

Remainders by the Binomial Trick

Base near a multiple of m

(km \pm 1)^n \equiv (\pm 1)^n \pmod{m}

Power of a Prime in n! (Legendre's Formula)

Legendre's formula

E_p(n!) = \sum_{i\ge 1} \left\lfloor \dfrac{n}{p^{\,i}} \right\rfloor

Common traps

Choose the base CLOSEST to a multiple of the divisor

Rewrite the base as the divisor (or a power of it)

\pm 1

.

8 = 7+1

for mod 7;

7 = 6+1

for mod 6/36. Picking a far-off form leaves many surviving terms and defeats the trick.

Count the prime, then divide by its exponent

For

8 = 2^3

, don't count "multiples of 8". Count the total power of 2 (via Legendre), then take the floor of that over 3. Counting 8s directly undercounts badly.

Binomial Theorem formulas

Coefficients & Specific Terms in the Expansion

The Binomial Theorem & the General Term

Binomial Coefficients — C(n, r)

Finding a Specific Term or Coefficient

The Middle Term

The Term Independent of x (Constant Term)

Conditions Linking Coefficients

Counting Terms in Products and Powers

Rational Terms & the General-Index Series

Common traps

Sums of Binomial Coefficients

Sum of All Coefficients — Put x = 1

Alternating & Odd/Even-Index Sums — Put x = −1

Weighted Sums via Differentiation

Pascal's Rule & Coefficient Identities

Common traps

Integer & Fractional Parts of Binomial Expressions

The Conjugate Trick — Irrational Parts Cancel

Integer Part + Fractional Part

Common traps

Remainders & Divisibility via Binomial Expansion

Remainders by the Binomial Trick

Power of a Prime in n! (Legendre's Formula)

Common traps

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