MathsCBSE Class 11 · Binomial Theorem

Sums of Binomial Coefficients, Remainders and Divisibility (Class 11)

Q: Why does the alternating sum equal zero?

Putting x = −1 in (1 + x)^n gives 0^n = 0 for n at least 1, and the left side is nC0 − nC1 + nC2 − …, so the alternating sum is 0.

Q: How is the sum of coefficients of a polynomial found?

Put x = 1. For example, the sum of coefficients of (3x − 2)^6 is (3 − 2)^6 = 1.

Q: Can the binomial theorem find any remainder?

It is best when some power of the base is one more or one less than a multiple of the divisor, such as 7² = 49 = 50 − 1 for divisor 25.

Q: Do these appear in board exams?

Yes. Coefficient sums, divisibility and remainder questions on the binomial theorem are standard NCERT-type board questions.

Updated 17 October 2026 · Fact-checked

Putting x = 1 in (1 + x)^n gives nC0 + nC1 + … + nCn = 2^n, and x = −1 gives the alternating sum 0, so the even-place and odd-place coefficients each add to 2^(n−1). For divisibility and remainders, write the base as (multiple of the divisor) ± 1 and expand with the binomial theorem; only the first few terms survive.

Put x = 1 for the total 2^n and x = −1 for 0; together they split the total equally, so even and odd places each give 2^(n−1).

Illustration drawn for QuestionBuddy.

Understand Sum of Binomial Coefficients and Divisibility Applications

The binomial theorem is also a tool for numbers. Two skills matter: summing coefficients by substituting special values, and splitting a large power into a multiple of the divisor plus a small leftover.

Sums of coefficients. From $(1 + x)^{n} = \sum^{n} C_{r} x^{r}$ :

$x = 1$ : $^{n} C_{0} +^{n} C_{1} + \dots +^{n} C_{n} = 2^{n}$ .
$x = - 1$ : $^{n} C_{0} -^{n} C_{1} +^{n} C_{2} - \dots = 0$ (for $n \geq 1$ ).
Adding and subtracting these two: $^{n} C_{0} +^{n} C_{2} +^{n} C_{4} + \dots =^{n} C_{1} +^{n} C_{3} + \dots = 2^{n - 1}$ .
For any polynomial $f (x)$ , the sum of its coefficients is $f (1)$ . The sum of coefficients of even powers is $\frac{f ( 1 ) + f ( - 1 )}{2}$ and of odd powers $\frac{f ( 1 ) - f ( - 1 )}{2}$ .
Using $r^{n} C_{r} = n^{n - 1} C_{r - 1}$ : $^{n} C_{1} + 2^{n} C_{2} + 3^{n} C_{3} + \dots + n^{n} C_{n} = n \cdot 2^{n - 1}$ .

Large powers. Write the base as round number $\pm$ small number. For example $10 1^{4} = (100 + 1)^{4}$ expands in four lines, and $(3 + 1)^{5} - (3 - 1)^{5}$ doubles the terms containing an odd power of 1 (the 2nd, 4th and 6th terms), because the terms containing an even power of 1 cancel.

Divisibility. To show $9^{n} - 8 n - 1$ is divisible by 64, write $9^{n} = (1 + 8)^{n} = 1 + 8 n + 64 (integer)$ , because every term from $8^{2}$ onwards contains $64$ . Hence $9^{n} - 8 n - 1 = 64 k$ . The general pattern: write the number as $(1 + d)^{n}$ where $d$ is a multiple of $m$ ; then $(1 + d)^{n} = 1 + n d + d^{2} (\dots)$ , so what remains after $1 + n d$ is divisible by $d^{2}$ .

Remainders. To find the remainder of $7^{103}$ on division by 25, note $7^{2} = 49 = 50 - 1$ , so $7^{103} = 7 (50 - 1)^{51}$ . Every term of $(50 - 1)^{51}$ except the last is a multiple of 25, so $(50 - 1)^{51} \equiv - 1$ and the remainder of $7^{103}$ is that of $- 7$ , namely $25 - 7 = 18$ . Always convert a negative leftover into a remainder between 0 and the divisor minus 1.

Last digits. The last two digits are the remainder on division by 100. Example: $7^{4} = 2401 = 2400 + 1$ , so $7^{100} = (2400 + 1)^{25}$ leaves remainder 1, last two digits 01.

Comparisons. For $x > 0$ , $(1 + x)^{n} > 1 + n x$ , and splitting $(100 \pm 1)^{n}$ shows which of two large expressions is bigger. Compare, never compute.

Sum of Binomial Coefficients and Divisibility Applications: formulas to remember

Sum of all coefficients: $n C 0 + n C 1 + \dots + n C n = 2^{n}$; Put x = 1 in (1 + x)^n.
Alternating sum: $n C 0 - n C 1 + n C 2 - \dots = 0$; Put x = −1; valid for n ≥ 1.
Even-place and odd-place sums: $n C 0 + n C 2 + \dots = n C 1 + n C 3 + \dots = 2^{(} n - 1)$; Each is half of 2^n.
Weighted sum: $1 \cdot n C 1 + 2 \cdot n C 2 + \dots + n \cdot n C n = n \times 2^{(} n - 1)$; Uses r·nCr = n·(n−1)C(r−1).
Divisibility pattern: $(1 + d)^{n} = 1 + n d + d^{2} \times (in t e g er)$; Everything after the second term is a multiple of d squared.

How to solve Sum of Binomial Coefficients and Divisibility Applications questions

Use these steps for coefficient sums, divisibility proofs, remainders and comparisons.

1For a sum of coefficients, decide which special value of $x$ ( $1$ , $- 1$ , or a combination) isolates the sum asked for.
2For a divisibility proof, rewrite the base as $1 + d$ or $d \pm 1$ so that $d$ contains the divisor.
3Expand using the binomial theorem and keep only the first two or three terms explicitly; group the rest with a common factor.
4Show that the leftover is a multiple of the divisor; for a remainder, convert a negative leftover into a value between 0 and the divisor minus 1.
5For a comparison, rewrite as a difference of two binomial expansions and compare the first surviving term.
6Check by substituting a small case such as $n = 1$ or $n = 2$ .

Sum of Binomial Coefficients and Divisibility Applications in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Counting subsets

Suppose a canteen offers 6 optional toppings for a sandwich, and you may take any number of them, including none.

How many different topping choices are there, and how many of them use an even number of toppings?

Try it first, then show the solutionSolution

Choosing $r$ toppings can be done in $^{6} C_{r}$ ways, so the total is $^{6} C_{0} +^{6} C_{1} + \dots +^{6} C_{6} = 2^{6} = 64$ .
Even numbers of toppings: $^{6} C_{0} +^{6} C_{2} +^{6} C_{4} +^{6} C_{6} = 1 + 15 + 15 + 1 = 32 = 2^{5}$ .

Answer: 64 choices in all; 32 use an even number of toppings.

Memory hook: Each topping is in or out: two options each, so 2 multiplied six times.

Remainders in a repeating schedule

Suppose a machine shows a code equal to the remainder when $3^{100}$ is divided by 7, and the technician wants it without a calculator.

Find the code.

Try it first, then show the solutionSolution

$3^{100} = 9^{50} = (7 + 2)^{50}$ , so the remainder equals that of $2^{50}$ .
$2^{50} = 4 \cdot 2^{48} = 4 \cdot 8^{16} = 4 (7 + 1)^{16}$ , whose remainder is that of $4 \cdot 1 = 4$ .
Check with the cycle $3^{6} \equiv 1$ : $3^{100} = 3^{96} \cdot 3^{4}$ , and $81 = 77 + 4$ leaves 4.

Answer: 4

Memory hook: Rewrite the base as a multiple of the divisor plus one: the 1 never changes.

Tricks to remember Sum of Binomial Coefficients and Divisibility Applications

Short memory aids. Say each one aloud once, then use it on a question the same day.

Everything is in or out
Each of n objects is either chosen or not, giving 2^n selections; this is why the coefficients add to 2^n.
Use it when: Recalling the coefficient sum instantly.
Round number plus one
To handle a^n mod m, hunt for a power of a that is one more or one less than a multiple of m.
Use it when: Remainder and last-digit questions.
Two terms, then a block
Write (1 + d)^n = 1 + nd + d²(…). The block is divisible by d².
Use it when: Divisibility by 64, 36, 25 and the like.
Negative leftover
A remainder of −7 on division by 25 is really 25 − 7 = 18.
Use it when: Final line of every remainder question.

Sum of Binomial Coefficients and Divisibility Applications: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A decimal power · 2 marks · Very short answerBoard level

Find $(1.02)^{5}$ correct to three decimal places using the binomial theorem.

Try it first, then show the solutionSolution

$(1 + 0.02)^{5} = 1 + 5 (0.02) + 10 (0.02)^{2} + 10 (0.02)^{3} + \dots$ . (1 mark)
$= 1 + 0.1 + 0.004 + 0.00008 + \dots = 1.10408 + (terms below 1 0^{- 6}) \approx 1.104$ . (1 mark)

Answer: 1.104

Example 2: Divisibility proof · 3 marks · Short answerBoard level

Prove that $1 6^{n} - 15 n - 1$ is divisible by 225 for every natural number $n$ .

Try it first, then show the solutionSolution

$1 6^{n} = (1 + 15)^{n} = 1 +^{n} C_{1} \cdot 15 +^{n} C_{2} \cdot 1 5^{2} + \dots +^{n} C_{n} 1 5^{n}$ . (1 mark)
$1 6^{n} - 15 n - 1 =^{n} C_{2} \cdot 1 5^{2} +^{n} C_{3} \cdot 1 5^{3} + \dots + 1 5^{n}$ . (1 mark)
Each term contains $1 5^{2} = 225$ , so the expression $= 225 \times$ an integer. (1 mark)

Answer: Divisible by 225.

Example 3: A remainder · 3 marks · Short answerHarder board level

Find the remainder when $7^{103}$ is divided by 25.

Try it first, then show the solutionSolution

$7^{103} = 7 \cdot (7^{2})^{51} = 7 \cdot 4 9^{51} = 7 (50 - 1)^{51}$ . (1 mark)
$(50 - 1)^{51} = 50 m - 1$ for an integer $m$ (all terms but the last have the factor 50). (1 mark)
$7 (50 m - 1) = 350 m - 7$ , and $350 m$ is a multiple of 25, so the remainder is that of $- 7$ , which is $25 - 7 = 18$ . (1 mark)

Answer: 18

Example 4: Even-and-odd split · 3 marks · Short answerHarder board level

Evaluate $(3 + 1)^{5} - (3 - 1)^{5}$ .

Try it first, then show the solutionSolution

Both expansions have the same terms, but signs differ on the terms with an odd power of $1$ or $- 1$ ; the difference doubles the terms where the power of $(\pm 1)$ is odd. (1 mark)
$= 2 [^{5} C_{1} (3)^{4} +^{5} C_{3} (3)^{2} +^{5} C_{5}]$ . (1 mark)
$= 2 [5 \cdot 9 + 10 \cdot 3 + 1] = 2 \cdot 76 = 152$ . (1 mark)

Answer: 152

Example 5: Comparing huge numbers · 5 marks · Long answerHOTS / competency

Show that $10 1^{50} > 10 0^{50} + 9 9^{50}$ .

Try it first, then show the solutionSolution

Rewrite as $10 1^{50} - 9 9^{50} > 10 0^{50}$ . (1 mark)
$10 1^{50} - 9 9^{50} = (100 + 1)^{50} - (100 - 1)^{50}$ . (1 mark)
Expansions subtract so that only odd powers of $1$ survive: $= 2 [^{50} C_{1} 10 0^{49} +^{50} C_{3} 10 0^{47} + \dots]$ . (1 mark)
The first term alone is $2 \cdot 50 \cdot 10 0^{49} = 10 0^{50}$ , and every other term is positive. (1 mark)
Hence $10 1^{50} - 9 9^{50} > 10 0^{50}$ , which gives $10 1^{50} > 10 0^{50} + 9 9^{50}$ . (1 mark)

Answer: Proved.

Where students lose marks in Sum of Binomial Coefficients and Divisibility Applications

The slips examiners see most often, and what to write instead.

Mistake: Writing 2^n for the even-place sum
The even-place and odd-place sums are each half of 2^n.
Wrong working
$^{8} C_{0} +^{8} C_{2} + \dots = 256$ .
Right working
$^{8} C_{0} +^{8} C_{2} + \dots = 2^{7} = 128$ .
Do this: Add the two sums: they must give 2^n.
Usually costs: The final mark.
Mistake: Leaving a negative remainder
A remainder must lie between 0 and divisor − 1.
Wrong working
Remainder of $7^{103}$ on division by 25 is $- 7$ .
Right working
$- 7 + 25 = 18$ .
Do this: Add the divisor once at the end.
Usually costs: 1 mark.
Mistake: Claiming divisibility without showing the factor
The examiner needs the common factor shown.
Wrong working
"Every term is divisible by 64, hence proved."
Right working
Show each remaining term contains $8^{2} = 64$ and write the expression as $64 k$ .
Do this: Write the final line as "= 64 × (integer)".
Usually costs: The last mark.
Mistake: Cancelling the wrong terms in (a+b)^n − (a−b)^n
The terms with an even power of b cancel, not the odd ones.
Wrong working
Keeping $^{5} C_{0} (3)^{5}$ in the difference.
Right working
The difference keeps $^{5} C_{1},^{5} C_{3},^{5} C_{5}$ terms (odd powers of 1).
Do this: Write the first two terms of each expansion with their signs.
Usually costs: 2 marks.

Exam tips

Put in the special value of $x$ explicitly: "putting $x = 1$ " and "putting $x = - 1$ " are the method marks.
For divisibility, always show the form $(1 + d)^{n}$ and the factor $d^{2}$ .
End a remainder question with a number between 0 and the divisor minus 1.
In comparison questions state the inequality you are proving in the first line, then rearrange it.
Test your final statement with $n = 1$ or $n = 2$ when it is a general result.

Practice questions on Sum of Binomial Coefficients and Divisibility Applications

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The value of $^{10} C_{0} +^{10} C_{2} +^{10} C_{4} + \dots +^{10} C_{10}$ is:
(a) 1024
(b) 512
(c) 1023
(d) 256
Hint
Even-place coefficients add to $2^{n - 1}$ .
Answer
(b). $2^{9} = 512$ .
Q2 · 1 mark · MCQHarder board level
The last two digits of $7^{100}$ are:
(a) 01
(b) 07
(c) 49
(d) 43
Hint
$7^{4} = 2401$ .
Answer
(a). $7^{100} = (2400 + 1)^{25}$ , and $2400$ is a multiple of 100, so the remainder on division by 100 is 1.
Q3 · 3 marks · Short answerHarder board level
Show that $3^{2 n + 2} - 8 n - 9$ is divisible by 64 for all natural numbers $n$ .
Hint
$3^{2 n + 2} = 9^{n + 1} = (1 + 8)^{n + 1}$ .
Answer
$9^{n + 1} = 1 + 8 (n + 1) + 64 (\dots)$ (terms from $8^{2}$ on). So $9^{n + 1} - 8 n - 9 = 1 + 8 n + 8 - 8 n - 9 + 64 (\dots) = 64 (\dots)$ , divisible by 64.
Q4 · 3 marks · Short answerHarder board level
Prove that $^{n} C_{1} + 2^{n} C_{2} + 3^{n} C_{3} + \dots + n^{n} C_{n} = n \cdot 2^{n - 1}$ and evaluate the sum for $n = 8$ .
Hint
Use $r^{n} C_{r} = n^{n - 1} C_{r - 1}$ .
Answer
$r^{n} C_{r} = n^{n - 1} C_{r - 1}$ , so the sum $= n (^{n - 1} C_{0} +^{n - 1} C_{1} + \dots +^{n - 1} C_{n - 1}) = n \cdot 2^{n - 1}$ . For $n = 8$ : $8 \cdot 128 = 1024$ .
Q5 · 5 marks · Long answerHOTS / competency
Show that the integer part of $(3 + 5)^{5}$ is 3935.
Hint
Add $(3 - 5)^{5}$ , a small positive number, to cancel the surds.
Answer
Let $S = (3 + 5)^{5} + (3 - 5)^{5} = 2 [3^{5} +^{5} C_{2} 3^{3} \cdot 5 +^{5} C_{4} 3 \cdot 25] = 2 [243 + 1350 + 375] = 3936$ . Since $0 < 3 - 5 < 1$ , $0 < (3 - 5)^{5} < 1$ . So $(3 + 5)^{5} = 3936 - (a number between 0 and 1)$ , whose integer part is 3935.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the sum nC0 + nC1 + … + nCn?Show answer
2^n.
2.What do the even-place coefficients add up to?Show answer
2^(n−1).
3.What is the sum of coefficients of a polynomial f(x)?Show answer
f(1).
4.What does r × nCr equal?Show answer
n × (n−1)C(r−1).
5.What is the first step in proving 16^n − 15n − 1 is divisible by 225?Show answer
Write 16^n = (1 + 15)^n and expand.

Sum of Binomial Coefficients and Divisibility Applications: frequently asked questions

Why does the alternating sum equal zero?

Putting x = −1 in (1 + x)^n gives 0^n = 0 for n at least 1, and the left side is nC0 − nC1 + nC2 − …, so the alternating sum is 0.

How is the sum of coefficients of a polynomial found?

Put x = 1. For example, the sum of coefficients of (3x − 2)^6 is (3 − 2)^6 = 1.

Can the binomial theorem find any remainder?

It is best when some power of the base is one more or one less than a multiple of the divisor, such as 7² = 49 = 50 − 1 for divisor 25.

Do these appear in board exams?

Yes. Coefficient sums, divisibility and remainder questions on the binomial theorem are standard NCERT-type board questions.