NDA Maths · Binomial Theorem

Sums of Binomial Coefficients

Sums of binomial coefficients are read off by substituting clever values of x into (1+x)ⁿ — x = 1 gives the total, x = −1 gives the alternating sum, and differentiating first gives the weighted sums.

All Binomial Theorem notes NDA Maths strategy

Why this matters

14 PYQs. The whole subtopic runs on one idea: the coefficients ARE the expansion, so plug a number into the right identity. x = 1, x = −1, and 'differentiate then substitute' cover almost everything; the Pascal-rule identities mop up the rest.

Concept 1 of 4

Sum of All Coefficients — Put x = 1

Intuition

The sum of all the coefficients of a polynomial is just its value at x = 1 — substituting 1 strips away every power of x and leaves the coefficients added up. For (1+x)ⁿ that total is 2ⁿ.

Definition

For any polynomial $f(x)$ , the sum of all coefficients is $f(1)$ . In particular:

$\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} = (1+1)^n = 2^n$ .
Dropping the first term: $\binom{n}{1} + \cdots + \binom{n}{n} = 2^n - 1$ .
A weighted base: $\sum_r \binom{n}{r} c^r = (1+c)^n$ (e.g. $\sum_r 2^r\binom{n}{r} = 3^n$ ).

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

Worked example

Find the sum of all coefficients in the expansion of

(2x + 3)^5

Sum of coefficients $= f(1) = (2\cdot 1 + 3)^5 = 5^5$ .

Answer:

3125

From the bank · past-year question

Example 1Binomial TheoremMODERATE

What is

C(n,1)+C(n,2)+\cdots+C(n,n)

equal to?

[Q46 · Apr · 2021]

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.

Drill 2 more on sum of all coefficients — put x = 1

Concept 2 of 4

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

Intuition

Substituting x = −1 flips the sign of every odd-power term, so it isolates the alternating sum — which collapses to zero for (1+x)ⁿ because the positives and negatives cancel exactly. Combining the x = 1 and x = −1 results splits the coefficients into odd-index and even-index halves.

Definition

Substitute $x = -1$ :

Alternating sum: $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \cdots = (1-1)^n = 0$ (for $n \ge 1$ ).
For a general polynomial, $a_0 - a_1 + a_2 - \cdots = f(-1)$ .
Odd/even split: sum of even-index coefficients $=$ sum of odd-index coefficients $= \dfrac{f(1)}{2} = 2^{\,n-1}$ . (They are equal because $f(-1) = 0$ .)
In $(a+b)^n + (a-b)^n$ , the odd-power terms cancel; in the difference, the even-power terms cancel.

Alternating sum and the split

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

Worked example

(1 - x + x^2)^4 = a_0 + a_1 x + \cdots + a_8 x^8

, find

a_0 - a_1 + a_2 - \cdots + a_8

The alternating sum is $f(-1)$ : substitute $x = -1$ .
$f(-1) = (1 - (-1) + (-1)^2)^4 = (1 + 1 + 1)^4 = 3^4$ .

Answer:

81

From the bank · past-year question

Example 2Binomial TheoremEASY

(1+2x-x^2)^6 = a_0 + a_1x + a_2x^2 + \ldots + a_{12}x^{12}

, then what is

a_0 - a_1 + a_2 - a_3 + a_4 - \ldots + a_{12}

equal to?

[Q3 · Apr · 2020]

Drill 3 more on alternating & odd/even-index sums — put x = −1

Concept 3 of 4

Weighted Sums via Differentiation

Intuition

A sum where each coefficient is multiplied by its index — like 1·C₁ + 2·C₂ + 3·C₃ + ⋯ — comes from DIFFERENTIATING (1+x)ⁿ once (which pulls a factor of the index down) and then substituting. Differentiate, then plug in x = 1 (or x = −1 for the alternating version).

Definition

Start from $(1+x)^n = \sum_r \binom{n}{r} x^r$ and differentiate:

n(1+x)^{n-1} = \sum_{r=1}^{n} r\binom{n}{r} x^{\,r-1}.

Put $x = 1$ : $\displaystyle\sum_{r=1}^{n} r\binom{n}{r} = n\,2^{\,n-1}$ .
Put $x = -1$ : $\displaystyle\sum_{r=1}^{n} (-1)^{r-1} r\binom{n}{r} = 0$ for $n > 1$ .

Multiplying by $x$ before differentiating, or differentiating twice, handles $\sum r^2\binom{n}{r}$ -type sums.

Index-weighted sum

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

Worked example

Evaluate

\,1\cdot\binom{n}{1} + 2\cdot\binom{n}{2} + \cdots + n\cdot\binom{n}{n}

This is $\sum_{r=1}^{n} r\binom{n}{r}$ .
Differentiate $(1+x)^n$ : $n(1+x)^{n-1} = \sum r\binom{n}{r}x^{r-1}$ , then set $x = 1$ .

Answer:

n\,2^{\,n-1}

From the bank · past-year question

Example 3Binomial TheoremMODERATE

Let

(1+x)^n = 1 + T_1 x + T_2 x^2 + \cdots + T_n x^n

What is

T_1+2T_2+3T_3+\cdots+nT_n

equal to?

[Q38 · Sep · 2023]

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.

Drill 2 more on weighted sums via differentiation

Concept 4 of 4

Pascal's Rule & Coefficient Identities

Intuition

A handful of structural identities — Pascal's rule, symmetry, and the alternating telescoping sum — collapse the trickiest coefficient questions to a single binomial coefficient. They are pattern-recognition tools: spot the shape, apply the identity.

Definition

The recurring identities:

Pascal's rule: $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$ .
Pascal applied twice: $\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2} = \binom{n+2}{r}$ .
Symmetry: $\binom{n}{r} = \binom{n}{n-r}$ , so the first and last coefficients are equal, and "coefficient of $a^m$ and $a^n$ in $(1+a)^{m+n}$ " are equal.
Middle-term split: $\binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n}$ (Pascal's rule on the central coefficient).

Pascal's rule (applied twice)

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

Worked example

Simplify

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

Group as $\left[\binom{n}{r} + \binom{n}{r-1}\right] + \left[\binom{n}{r-1} + \binom{n}{r-2}\right]$ .
Each bracket is Pascal's rule: $\binom{n+1}{r} + \binom{n+1}{r-1}$ .
Apply Pascal's rule once more.

Answer:

\binom{n+2}{r}

From the bank · past-year question

Example 4Binomial TheoremHARD

What is

C(n,r) + 2C(n,r-1) + C(n,r-2)

equal to?

[Q28 · Apr · 2018]

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.

Drill 3 more on pascal's rule & coefficient identities

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

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

Watch out for (3)

Sum of coefficients uses x = 1, not x = 0
→ Sum of All Coefficients — Put x = 1
Differentiate first, substitute second
→ Weighted Sums via Differentiation
Pascal's rule needs adjacent lower indices on the SAME n
→ Pascal's Rule & Coefficient Identities

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Binomial TheoremEASY

C_0, C_1, C_2, \ldots, C_n

are the coefficients in the expansion of

(1+x)^n

, then what is the value of

C_1 + C_2 + C_3 + \cdots + C_n

[Q4 · Apr · 2021]

Example 2Binomial TheoremMODERATE

How many terms are there in the expansion of

(1 + 2x + x^{2})^{5} + (1 + 4y + 4y^{2})^{5}

[Q24 · Sep · 2019]

Example 3Binomial TheoremMODERATE

Let

(1+x)^n = 1 + T_1 x + T_2 x^2 + \cdots + T_n x^n

What is

1-T_1+2T_2-3T_3+\cdots+(-1)^n nT_n

equal to?

[Q39 · Sep · 2023]

Example 4Binomial TheoremEASY

If the coefficients of

a^m

and

a^n

in the expansion of

(1+a)^{m+n}

are

\alpha

and

\beta

, then which one is correct?

[Q10 · Apr · 2018]

Example 5Binomial TheoremEASY

What is

\sum_{r=0}^{n} 2^r C(n,r)

equal to?

[Q2 · Sep · 2022]

Drill every past-year question on this subtopic

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