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7.2.1 Some Special Cases

🎓 Class 11 Mathematics CBSE Theory Ch 7 — Binomial Theorem ⏱ ~15 min
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7.2.1 Some Special Cases

Substituting specific values of \(a\) and \(b\) in the theorem produces several frequently-used identities.

Special Cases
(i) Taking \(a=1,\ b=x\): \[(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\cdots+\binom{n}{n}x^n.\] (ii) Taking \(a=1,\ b=-x\): \[(1-x)^n=\binom{n}{0}-\binom{n}{1}x+\binom{n}{2}x^2-\cdots+(-1)^n\binom{n}{n}x^n.\] (iii) Taking \(x=1\) in (i): \(\sum_{k=0}^{n}\binom{n}{k}=2^n\) (row sum).
(iv) Taking \(x=1\) in (ii): \(\sum_{k=0}^{n}(-1)^k\binom{n}{k}=0\) (alternating row sum, for \(n\ge 1\)).

7.2.2 General Term and Middle Term

General term
The \((k+1)\)th term in the expansion of \((a+b)^n\) is \[\boxed{\;T_{k+1}=\binom{n}{k}a^{n-k}b^{k}\;}\] Counting starts from \(k=0\), so \(T_1=\binom{n}{0}a^n\), \(T_2=\binom{n}{1}a^{n-1}b\), and so on up to \(T_{n+1}=\binom{n}{n}b^n\).
Middle term
There are \(n+1\) terms; the middle position depends on parity of \(n\):
  • n even: single middle term \(T_{(n/2)+1}=\binom{n}{n/2}\,a^{n/2}b^{n/2}\).
  • n odd: two middle terms \(T_{(n+1)/2}\) and \(T_{(n+3)/2}\).

Worked Examples

Example 3. Which is larger \((1.01)^{1\,000\,000}\) or \(10\,000\)?
\((1.01)^{1\,000\,000}=(1+0.01)^{1\,000\,000}=\binom{10^6}{0}+\binom{10^6}{1}(0.01)+\text{positive terms}\) \(=1+10^6\cdot 0.01+\ldots=1+10\,000+\ldots>10\,000\). So \((1.01)^{10^6}>10\,000\).
Example 4. Use the binomial theorem to prove that \(6^n-5n\) always leaves remainder 1 when divided by 25.
\(6^n=(1+5)^n=\binom{n}{0}+\binom{n}{1}\cdot 5+\binom{n}{2}\cdot 5^2+\cdots+\binom{n}{n}\cdot 5^n\) \(=1+5n+25\!\left[\binom{n}{2}+\binom{n}{3}\cdot 5+\cdots+\binom{n}{n}\cdot 5^{n-2}\right]=1+5n+25k\) for some integer \(k\).
Hence \(6^n-5n=1+25k\), i.e. \(6^n-5n\equiv 1\pmod{25}\). \(\square\)
Exercise 7.1 — Expand using Binomial Theorem
1. Expand \((1-2x)^5\).
\(\binom{5}{0}-\binom{5}{1}(2x)+\binom{5}{2}(2x)^2-\binom{5}{3}(2x)^3+\binom{5}{4}(2x)^4-\binom{5}{5}(2x)^5\) \(=1-10x+40x^2-80x^3+80x^4-32x^5\).
2. Expand \(\left(\dfrac{2}{x}-\dfrac{x}{2}\right)^5\).
With \(a=2/x, b=-x/2\): \(T_{k+1}=\binom{5}{k}(2/x)^{5-k}(-x/2)^k\). Computing all 6 terms: \(\dfrac{32}{x^5}-\dfrac{40}{x^3}+\dfrac{20}{x}-5x+\dfrac{5x^3}{8}-\dfrac{x^5}{32}\).
3. Expand \((2x-3)^6\).
Coefficients of row 6: 1, 6, 15, 20, 15, 6, 1. Powers of 2x decrease 6→0; powers of (−3) increase 0→6. Computing: \(64x^6-576x^5+2160x^4-4320x^3+4860x^2-2916x+729\).
4. Expand \(\left(\dfrac{x}{3}+\dfrac{1}{x}\right)^5\).
\(T_{k+1}=\binom{5}{k}(x/3)^{5-k}(1/x)^k=\binom{5}{k}\dfrac{x^{5-2k}}{3^{5-k}}\). Computing for \(k=0,\ldots,5\): \(\dfrac{x^5}{243}+\dfrac{5x^3}{81}+\dfrac{10x}{27}+\dfrac{10}{9x}+\dfrac{5}{3x^3}+\dfrac{1}{x^5}\).
5. Expand \(\left(x+\dfrac{1}{x}\right)^6\).
Row 6 coefs (1,6,15,20,15,6,1) and powers \(x^{6-2k}\): \(x^6+6x^4+15x^2+20+\dfrac{15}{x^2}+\dfrac{6}{x^4}+\dfrac{1}{x^6}\).
6. Use binomial theorem to evaluate \((96)^3\).
\((100-4)^3=100^3-3\cdot 100^2\cdot 4+3\cdot 100\cdot 16-64=10^6-1.2\!\cdot\!10^5+4800-64=884\,736\).
7. \((102)^5\).
\((100+2)^5=100^5+5\cdot 100^4\cdot 2+10\cdot 100^3\cdot 4+10\cdot 100^2\cdot 8+5\cdot 100\cdot 16+32\) \(=10^{10}+10^9+4\!\cdot\!10^7+8\!\cdot\!10^5+8000+32=11\,040\,808\,032\).
8. \((101)^4\).
\((100+1)^4=100^4+4\cdot 100^3+6\cdot 100^2+4\cdot 100+1=10^8+4\!\cdot\!10^6+6\!\cdot\!10^4+400+1=104\,060\,401\).
9. \((99)^5\).
\((100-1)^5=10^{10}-5\!\cdot\!10^8+10\!\cdot\!10^6-10\!\cdot\!10^4+5\!\cdot\!100-1=9\,509\,900\,499\).
10. Using binomial theorem, indicate which number is larger \((1.1)^{10\,000}\) or \(1000\)?
\((1.1)^{10000}=(1+0.1)^{10000}=1+10000(0.1)+\text{positive}=1+1000+\text{positive}>1000\). So \((1.1)^{10\,000}>1000\).
11. Find \((a+b)^4-(a-b)^4\), and hence evaluate \((\sqrt 3+\sqrt 2)^4-(\sqrt 3-\sqrt 2)^4\).
Subtraction kills even-k terms (those without sign change), keeps odd-k terms doubled: \((a+b)^4-(a-b)^4=2[\binom{4}{1}a^3b+\binom{4}{3}ab^3]=2[4a^3b+4ab^3]=8ab(a^2+b^2)\).
With \(a=\sqrt 3, b=\sqrt 2\): \(=8\sqrt 6\cdot 5=40\sqrt 6\).
12. Find \((x+1)^6+(x-1)^6\), and hence evaluate \((\sqrt 2+1)^6+(\sqrt 2-1)^6\).
Addition keeps even-k, kills odd-k. Result \(=2[\binom{6}{0}x^6+\binom{6}{2}x^4+\binom{6}{4}x^2+\binom{6}{6}]=2[x^6+15x^4+15x^2+1]\).
With \(x=\sqrt 2\): \(=2[8+15\cdot 4+15\cdot 2+1]=2[8+60+30+1]=2\cdot 99=198\).
13. Show that \(9^{n+1}-8n-9\) is divisible by 64.
\(9^{n+1}=(1+8)^{n+1}=1+(n+1)\cdot 8+\binom{n+1}{2}\cdot 64+\binom{n+1}{3}\cdot 8^3+\cdots\). So \(9^{n+1}-8n-9=1+8n+8-8n-9+64\!\cdot\!\binom{n+1}{2}+64\cdot 8\!\cdot\!\binom{n+1}{3}+\cdots=64\cdot[\text{integer}]\), divisible by 64. \(\square\)
14. Prove \(\sum_{r=0}^{n}3^r\binom{n}{r}=4^n\).
This is just \((1+3)^n=4^n\) expanded by the binomial theorem with \(a=1, b=3\).
Miscellaneous Exercise on Chapter 7
1. If \(a, b\) are distinct integers, prove that \(a-b\) is a factor of \(a^n-b^n\), \(n\) positive integer.
Write \(a=(a-b)+b\). Then \(a^n=[(a-b)+b]^n=\sum_{k=0}^{n}\binom{n}{k}(a-b)^{n-k}b^k\). The last term (k=n) is \(b^n\); subtract: \(a^n-b^n=\sum_{k=0}^{n-1}\binom{n}{k}(a-b)^{n-k}b^k=(a-b)\sum_{k=0}^{n-1}\binom{n}{k}(a-b)^{n-k-1}b^k\). The bracket is an integer, so \(a-b\) divides \(a^n-b^n\). \(\square\)
2. Evaluate \((\sqrt 3+\sqrt 2)^6-(\sqrt 3-\sqrt 2)^6\).
Using the same idea as Q11 of Exercise 7.1 with \(n=6\): subtraction keeps only odd-\(k\) terms doubled. \((\sqrt 3+\sqrt 2)^6-(\sqrt 3-\sqrt 2)^6=2[\binom{6}{1}(\sqrt 3)^5(\sqrt 2)+\binom{6}{3}(\sqrt 3)^3(\sqrt 2)^3+\binom{6}{5}(\sqrt 3)(\sqrt 2)^5]\) \(=2[6\cdot 9\sqrt 3\cdot\sqrt 2+20\cdot 3\sqrt 3\cdot 2\sqrt 2+6\sqrt 3\cdot 4\sqrt 2]\) \(=2[54\sqrt 6+120\sqrt 6+24\sqrt 6]=2\cdot 198\sqrt 6=396\sqrt 6\).
3. Find the value of \((a^2+\sqrt{a^2-1})^4+(a^2-\sqrt{a^2-1})^4\).
Let \(p=a^2,\ q=\sqrt{a^2-1}\). Sum keeps even-\(k\) doubled. \(2[\binom{4}{0}p^4+\binom{4}{2}p^2q^2+\binom{4}{4}q^4]=2[a^8+6a^4(a^2-1)+(a^2-1)^2]\) \(=2[a^8+6a^6-6a^4+a^4-2a^2+1]=2a^8+12a^6-10a^4-4a^2+2\).
4. Find an approximation of \((0.99)^5\) using the first three terms.
\((1-0.01)^5\approx 1-5(0.01)+10(0.01)^2=1-0.05+0.001=0.951\). (Actual: 0.9509900499.)
5. Expand using Binomial Theorem \(\left(1+\dfrac{x}{2}-\dfrac{2}{x}\right)^4,\ x\ne 0\).
Group as \([1+(\tfrac{x}{2}-\tfrac{2}{x})]^4=\sum_{k=0}^{4}\binom{4}{k}(\tfrac{x}{2}-\tfrac{2}{x})^k\). Compute term-by-term: it works out to \(1+2x-\dfrac{8}{x}+\dfrac{3x^2}{2}+\dfrac{24}{x^2}-12+\dfrac{x^3}{2}-\dfrac{32}{x^3}+\dfrac{x^4}{16}+\dfrac{16}{x^4}-3x+\dfrac{12}{x}\) — terms collected by power of \(x\). (Long but routine.)
6. Find the expansion of \((3x^2-2ax+3a^2)^3\) using the binomial theorem.
Write \((3x^2-2ax+3a^2)=[(3x^2+3a^2)-2ax]\). Then cube using \(n=3\): \([(A)-2ax]^3=A^3-3A^2(2ax)+3A(2ax)^2-(2ax)^3\) where \(A=3x^2+3a^2=3(x^2+a^2)\). \(A^3=27(x^2+a^2)^3=27[x^6+3x^4a^2+3x^2a^4+a^6]\); etc. Result simplifies to \(27x^6-54ax^5+117a^2x^4-116a^3x^3+117a^4x^2-54a^5x+27a^6\).
Activity: Spot the Term Independent of x
L4 Analyse
Materials: Pen, paper.
Predict: In the expansion of \((x+1/x)^6\), is there a constant term (no x)? At what \(k\)?
  1. Write the general term \(T_{k+1}=\binom{6}{k}x^{6-k}(1/x)^k=\binom{6}{k}x^{6-2k}\).
  2. For the term to be independent of \(x\), set \(6-2k=0\), giving \(k=3\).
  3. So the constant term is \(T_4=\binom{6}{3}=20\).
  4. Now try \((x^2-1/x)^9\). Find the term independent of \(x\). (Hint: power of \(x\) in \(T_{k+1}\) is \(2(9-k)-k=18-3k\); set to 0.)
For \((x^2-1/x)^9\): \(T_{k+1}=\binom{9}{k}(x^2)^{9-k}(-1/x)^k=\binom{9}{k}(-1)^k x^{18-3k}\). Set \(18-3k=0\Rightarrow k=6\). So term independent of \(x\) is \(\binom{9}{6}(-1)^6=84\). The "find power = 0" technique works for any expansion of this form.

Consolidation Competency-Based Questions

Scenario: A statistician analysing a coin toss problem expands \((p+q)^{10}\) where \(p\) is "heads" probability and \(q=1-p\). The coefficient \(\binom{10}{k}p^{10-k}q^k\) gives the probability of exactly \(k\) tails in 10 tosses.
Q1. The middle term in the expansion of \((1+x)^{10}\) is:
L3 Apply
  • (a) \(\binom{10}{4}x^4\)
  • (b) \(\binom{10}{5}x^5\)
  • (c) \(\binom{10}{6}x^6\)
  • (d) \(\binom{10}{5}x^4\)
Answer: (b). n=10 is even; middle term is \(T_{6}=\binom{10}{5}x^5=252\,x^5\).
Q2. (Fill in) The general term in \((1+x)^n\) is \(T_{k+1}=\) ____ . The coefficient of \(x^7\) in \((1+x)^{12}\) is ____ .
L2 Understand
Answer: \(\binom{n}{k}x^k\); coefficient of \(x^7\) is \(\binom{12}{7}=792\).
Q3. (T/F) "The coefficient of \(x^k\) in \((1+x)^n\) equals the coefficient of \(x^{n-k}\) in the same expansion." Justify.
L5 Evaluate
True. By symmetry of binomial coefficients \(\binom{n}{k}=\binom{n}{n-k}\). Geometrically: rows of Pascal's Triangle are palindromes.
Q4. Apply: in the coin-toss scenario, the probability of exactly 6 tails in 10 fair tosses is:
L4 Analyse
Answer: \(p=q=1/2\). \(\binom{10}{6}(0.5)^4(0.5)^6=210/1024\approx 0.205\) or about 20.5%.
Q5. Design: prove \(\binom{n}{0}^2+\binom{n}{1}^2+\cdots+\binom{n}{n}^2=\binom{2n}{n}\) using the binomial theorem applied to \((1+x)^n(1+x)^n=(1+x)^{2n}\) and matching the coefficient of \(x^n\).
L6 Create
Solution: LHS coefficient of \(x^n\) in \((1+x)^n(1+x)^n\): pick \(x^k\) from first factor and \(x^{n-k}\) from second, sum over \(k\): \(\sum_k\binom{n}{k}\binom{n}{n-k}=\sum_k\binom{n}{k}^2\) (using \(\binom{n}{n-k}=\binom{n}{k}\)). RHS coefficient of \(x^n\) in \((1+x)^{2n}\) is \(\binom{2n}{n}\). Equate. \(\square\)

Consolidation Assertion–Reason

Assertion (A): \(6^n-5n\) leaves remainder 1 when divided by 25.
Reason (R): \(6^n=(1+5)^n=1+5n+25k\) for some integer \(k\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). Subtracting \(5n\): \(6^n-5n=1+25k\). R gives A directly via the binomial expansion.
Assertion (A): The middle term of \((1+x)^7\) is uniquely defined.
Reason (R): When \(n\) is odd, the expansion has \(n+1\) (even) terms, so there is no single middle term but two.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (d). A is false — for n=7, there are 8 terms; the two middle ones are \(T_4\) and \(T_5\). R is true and is the precise reason A is false.

Chapter Summary

Key formulas
  • Binomial Theorem: \((a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k\) for positive integer \(n\).
  • Binomial coefficient: \(\binom{n}{k}=\dfrac{n!}{k!(n-k)!}\); symmetric \(\binom{n}{k}=\binom{n}{n-k}\); Pascal \(\binom{n}{k}+\binom{n}{k-1}=\binom{n+1}{k}\).
  • Number of terms: \(n+1\). General term \(T_{k+1}=\binom{n}{k}a^{n-k}b^k\).
  • Middle term: if \(n\) even, \(T_{n/2+1}\); if \(n\) odd, \(T_{(n+1)/2}\) and \(T_{(n+3)/2}\).
  • Special cases: \((1+x)^n,\ (1-x)^n\); \(\sum\binom{n}{k}=2^n\); \(\sum(-1)^k\binom{n}{k}=0\) for \(n\ge 1\).
  • Term independent of x: set the exponent of \(x\) in \(T_{k+1}\) to 0 and solve for \(k\).

Historical Note

The ancient Indian mathematicians knew about the coefficients in the expansions of \((x+y)^n,\ 0\le n\le 7\). The arrangement of these coefficients in the form of a diagram called Meru-Prastara, provided by Pingala in his book Chhandahsastra (200 BC), is a kind of arithmetic triangle and is the same as the Pascal's triangle. Halayudha (10th century) commented on Pingala's text and noted explicit formulas for these coefficients. Bhaskaracharya II, in his work Lilavati (1150), gives the rule for finding binomial coefficients.

Outside India, the arithmetic triangle was known to the Persian mathematician Al-Karaji (c. 1000), the Chinese mathematician Yang Hui (1261), and several others. The present form of the binomial theorem for integer values of \(n\) appeared in Pascal's Traité du Triangle Arithmétique, published posthumously in 1665. Newton's generalisation to non-integer exponents (1665) gave us the binomial series and the modern theory of power-series expansions.

Frequently Asked Questions

What is the general term in the binomial expansion?
The (k + 1)th term in the expansion of (a + b)ⁿ is Tₖ₊₁ = ⁿCₖ aⁿ⁻ᵏ bᵏ. Setting different values of k gives every term in the expansion.
What is the middle term in (a + b)ⁿ?
If n is even, there is one middle term: T_{(n/2)+1} = ⁿC_{n/2} aⁿ/² bⁿ/². If n is odd, there are two middle terms: T_{(n+1)/2} and T_{(n+3)/2}.
How do you find the term independent of x?
In a binomial expansion involving x, write the general term Tₖ₊₁ and simplify the power of x in it. Set that exponent equal to 0 and solve for k. The corresponding term is the term independent of x (the constant term).
What is the binomial theorem useful for?
Computing approximate powers (e.g. (1.001)^100), verifying divisibility (e.g. proving 6ⁿ − 5n leaves remainder 1 when divided by 25), expanding expressions in power-series form, and as a stepping stone to the Newton generalisation used throughout calculus and statistics.
What is the sum 1 + ⁿC₁ + ⁿC₂ + ⋯ + ⁿCₙ?
Setting a = b = 1 in (a + b)ⁿ gives 2ⁿ. So the row sum of Pascal's Triangle for row n is exactly 2ⁿ.
Why does (1 − x)ⁿ have alternating signs?
Substituting b = −x in the theorem multiplies each term by (−1)ᵏ, where k is the power of x. So the sign alternates with the power of x.
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