K2KARAKORAM
मराठीEnglish
Open the app

State Board · Class 11 · Mathematics Part 2

Lesson 4: Induction and Binomial Theorem

5 concepts · 40 practice questions with answers · Free to read

This lesson is also in: मराठी

Principle of mathematical induction

From daily life

Dominoes stand close in a row: if the first falls and every falling domino knocks down the next, all of them fall. Proving a statement true for every natural number works the same way. Building on classes 9–10 : Since classes 9–10 we have checked formulas on examples; here we prove a formula for every n.

Simple definition

For a statement P(n), if (1) P(1) is true and (2) P(k) true implies P(k + 1) true, then P(n) is true for every natural number n.

Rule 1: Four steps — Foundation: check n = 1. Assumption: take it true for n = k. Succession: prove it for n = k + 1. Conclusion: true for all n.

P(1) true + P(k) gives P(k+1)12345k1st falls → kth falls → (k+1)th falls

In the figure : Dominoes: the first falls and each knocks down the next, so all fall.

P(1)P(2)P(3)P(4)baseP(k) ⇒ P(k+1)Step 1: check n = 1Step 2: assume n = k, prove n = k + 1Step 3: conclude for all n

In the figure : Chain: P(1) is true, then P(1) gives P(2), P(2) gives P(3) and so on.

How is the formula obtained? : P(1) is true. By step (2), P(1) gives P(2), P(2) gives P(3) and so on without end, so every natural number lies on this chain.

  • Easy : 1 + 2 + 3 + … + n = n(n + 1)/2. For n = 1: left side 1, right side 1 × 2/2 = 1. Assuming it for k and adding (k + 1): k(k + 1)/2 + (k + 1) = (k + 1)(k + 2)/2.
  • Medium : 1 + 3 + 5 + … + (2n − 1) = n². Assuming it for k and adding (2k + 1): k² + 2k + 1 = (k + 1)².
  • Exam-type : 1² + 2² + … + n² = n(n + 1)(2n + 1)/6. Succession: k(k + 1)(2k + 1)/6 + (k + 1)² = (k + 1)(2k² + 7k + 6)/6 = (k + 1)(k + 2)(2k + 3)/6.
Remember

The succession step must use the assumption; a check that does not use it is not a proof.

Common mistake

Stopping after checking n = 1, 2, 3. Induction needs n = k to give n = k + 1.

Shortcut

In the succession step always write "sum of k terms + the (k + 1)-th term" and compare with the right side at n = k + 1. Memory clue : "Stand on k, jump to k + 1".

CET trap

When writing P(k + 1) put k + 1 in place of n everywhere, not at just one place.

CET-style example : By 1 + 2 + … + n = n(n + 1)/2, the sum for n = 20 is? (A) 190 (B) 210 (C) 200 (D) 220

Answer : (B). 20 × 21/2 = 210. Time-saving method : Apply the formula directly; 19 × 20/2 = 190 is for n = 19.

Spot the difference

PointCheckingProof by induction
Values of na few chosen nall n
Methodput values and seefoundation + succession
Conclusiononly for those nfor all natural numbers

Practice questions

Easy
  1. 1For 2 + 4 + … + 2n = n(n + 1), what is each side at n = 1?
  2. 2Find the sum of natural numbers from 1 to 6 by the formula.
  3. 3Find the sum of the first 10 odd numbers.
Medium
  1. 4If 1 + 2 + … + k = k(k + 1)/2, what is the formula for 1 + 2 + … + (k + 1)?
  2. 5Find 1² + 2² + … + 10².
  3. 6Find the sum of the first 20 even numbers.
Challenging
  1. 7Using 1·2 + 2·3 + … + n(n + 1) = n(n + 1)(n + 2)/3 find the sum for n = 10.
  2. 8Using 1/(1·2) + 1/(2·3) + … + 1/(n(n + 1)) = n/(n + 1) find the sum for n = 9.
Show answers (try them yourself first)
  1. 1दोन्ही बाजू 2.
  2. 26 × 7/2 = 21.
  3. 310² = 100.
  4. 4(k + 1)(k + 2)/2.
  5. 510 × 11 × 21/6 = 385.
  6. 620 × 21 = 420.
  7. 710 × 11 × 12/3 = 440.
  8. 89/10.

Proving divisibility by induction

From daily life

We cannot test one n at a time to see whether an expression is always divisible by 6. Induction does it once for all. Building on classes 9–10 : Since classes 9–10 we know products and factors; here we split into factors to show divisibility.

Simple definition

In a divisibility statement P(k) means "expression = m × the divisor". In P(k + 1) separate the P(k) part and show that the remaining part also has the same divisor.

Rule 1: The splitting trick — aᵏ⁺¹ − 1 = a(aᵏ − 1) + (a − 1). (k + 1)³ = k³ + 3k² + 3k + 1. Factorise the leftover part until the divisor shows.

P(1)P(2)P(3)P(4)baseP(k) ⇒ P(k+1)Step 1: check n = 1Step 2: assume n = k, prove n = k + 1Step 3: conclude for all n

In the figure : Chain: once P(1) is true and P(k) gives P(k + 1), divisibility holds for all n.

How is the formula obtained? : 7ᵏ⁺¹ − 1 = 7(7ᵏ − 1) + 6. By P(k) the number 7ᵏ − 1 is divisible by 6, and so is 6; hence so is the sum.

  • Easy : 7ⁿ − 1 is divisible by 6. n = 1: 6. 7ᵏ⁺¹ − 1 = 7(7ᵏ − 1) + 6; both terms are divisible by 6.
  • Medium : n³ − n is divisible by 6. (k + 1)³ − (k + 1) = (k³ − k) + 3k(k + 1). Since k(k + 1) is even, 3k(k + 1) is divisible by 6.
  • Exam-type : 9ⁿ − 1 is divisible by 8. 9ᵏ⁺¹ − 1 = 9(9ᵏ − 1) + 8; n = 1: 8. Also 4ⁿ + 15n − 1 is divisible by 9 (n = 1: 18).
Remember

aⁿ − bⁿ is always divisible by (a − b); hence 7ⁿ − 1 by 6 and 9ⁿ − 1 by 8.

Common mistake

Rewriting only the n = 1 check in the succession step. Show the term obtained from the assumption.

Shortcut

Split into two parts: (the P(k) part) + (the leftover); show the leftover has the divisor. Memory clue : "Old part + leftover part".

CET trap

If the value at n = 1 is not divisible by the divisor the statement is false; check this first.

CET-style example : The number that always divides 7ⁿ − 1 is? (A) 6 (B) 8 (C) 12 (D) 9

Answer : (A). n = 1: 6; n = 2: 48; the one dividing both is 6. Time-saving method : The value at n = 1 is 6; test the options against it.

Spot the difference

ExpressionDivisorValue at n = 1
7ⁿ − 166
9ⁿ − 188
n³ − n60

Practice questions

Easy
  1. 1Check for n = 1 that 5ⁿ − 1 is divisible by 4.
  2. 2Is 5⁴ − 1 divisible by 4?
  3. 3For n = 5 which is larger, 2ⁿ or n?
Medium
  1. 4Check that n³ + 2n is divisible by 3 for n = 1 to 5 and write the conclusion.
  2. 5Find (8³ − 1)/7.
  3. 6By which number is n(n + 1)(n + 2) always divisible?
Challenging
  1. 7Check for n = 1 that 10ⁿ + 3·4ⁿ⁺² + 5 is divisible by 9.
  2. 8Check for n = 2 that 2³ⁿ − 1 is divisible by 7.
Show answers (try them yourself first)
  1. 15 − 1 = 4 ; 4 ने भाग जातो.
  2. 2624 = 4 × 156 ; हो.
  3. 32⁵ = 32 > 5.
  4. 4n = 1 : 3, 2 : 12, 3 : 33, 4 : 72, 5 : 135 ; सर्वांना 3 ने भाग जातो.
  5. 5511/7 = 73.
  6. 66 ने.
  7. 710 + 192 + 5 = 207 = 9 × 23.
  8. 82⁶ − 1 = 63 = 7 × 9.

Binomial theorem

From daily life

We expanded the square and cube of (a + b) in class 9. For power 10 repeated multiplication is hard; the binomial theorem gives the terms directly. Building on classes 9–10 : Since classes 9–10 we know (a + b)² and (a + b)³; here is the formula for any positive integer power.

Simple definition

For a positive integer n, (a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ aⁿ⁻¹ b + ⁿC₂ aⁿ⁻² b² + … + ⁿCₙ bⁿ. There are n + 1 terms and in each term the powers of a and b add up to n.

Rule 1: Coefficients — The coefficients come from the n-th row of Pascal’s triangle: 1, 4, 6, 4, 1 (n = 4). For (a − b)ⁿ the signs alternate + − + −.

101111212133131464141510105154+6 = 10row n gives coefficients of (a+b)ⁿ

In the figure : Pascal’s triangle: each number is the sum of the two above it.

How is the formula obtained? : When (a + b)ᵏ is multiplied by (a + b), the terms ᵏCᵣ₋₁ and ᵏCᵣ meet at the same power and add to ᵏ⁺¹Cᵣ; so the formula passes from k to k + 1.

  • Easy : (x + 2)⁴ = x⁴ + 8x³ + 24x² + 32x + 16.
  • Medium : (2x − 1)⁵ = 32x⁵ − 80x⁴ + 80x³ − 40x² + 10x − 1. Coefficients 1, 5, 10, 10, 5, 1, each multiplied by powers of 2x and of −1.
  • Exam-type : 101³ = (100 + 1)³ = 1000000 + 30000 + 300 + 1 = 1030301. (1.02)⁴ = 1 + 0.08 + 0.0024 + 0.000032 + … ≈ 1.0824 (four decimals).
Remember

There are n + 1 terms; the coefficients of (a + b)ⁿ add up to 2ⁿ.

Common mistake

Forgetting the sign of b in (a − b)ⁿ. Write it as (a + (−b))ⁿ.

Shortcut

A large power: write the number as (big + small), e.g. 101 = 100 + 1; 998 = 1000 − 2. Memory clue : "Round number + a little".

CET trap

Powers of x and of the multiplier come together: (2x)³ = 8x³, not 2x³.

CET-style example : The sum of all coefficients of (1 + x)¹⁰ is? (A) 1024 (B) 100 (C) 512 (D) 2048

Answer : (A). Put x = 1: 2¹⁰ = 1024. Time-saving method : For the sum of all coefficients put x = 1; do not expand.

Spot the difference

Point(a + b)ⁿ(a − b)ⁿ
Termsn + 1n + 1
Signsall +alternately +, −
Sum of coefficients2ⁿ0 (for a = b = 1)

Practice questions

Easy
  1. 1Expand (x + 3)³.
  2. 2Write the row of coefficients of (a + b)⁵.
  3. 3What is 99²?
Medium
  1. 4Expand (x − 1)⁴.
  2. 5Find the sum of the coefficients of (1 + x)⁶.
  3. 6Find 102³.
Challenging
  1. 7Find the sum of the coefficients of (1 − x)⁸ with their signs.
  2. 8Find (0.99)⁵ correct to three decimals.
Show answers (try them yourself first)
  1. 1x³ + 9x² + 27x + 27.
  2. 21, 5, 10, 10, 5, 1.
  3. 3(100 − 1)² = 10000 − 200 + 1 = 9801.
  4. 4x⁴ − 4x³ + 6x² − 4x + 1.
  5. 52⁶ = 64.
  6. 6(100 + 2)³ = 1000000 + 60000 + 1200 + 8 = 1061208.
  7. 7x = 1 घातला : 0.
  8. 8(1 − 0.01)⁵ = 1 − 0.05 + 0.001 − … ≈ 0.951.

General term and middle term

From daily life

To get the fifth term of (a + b)¹⁵ there is no need to write the whole expansion. A formula for one term gives it directly. Building on classes 9–10 : Since classes 9–10 we know the laws of indices; here we use them to fix the powers in a term.

Simple definition

The (r + 1)-th term of (a + b)ⁿ is Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ. The middle term: for even n there is one, T(n/2 + 1); for odd n there are two, T((n + 1)/2) and T((n + 3)/2).

Rule 1: Finding a particular term — Find r by equating the power of x in Tᵣ₊₁ to the required power. The term independent of x means the power of x is 0.

1T16T215T320T415T56T61T7n = 6 even: one middle term T4T(r+1) = ⁿCᵣ aⁿ⁻ʳ bʳ

In the figure : n = 6: seven terms; the middle one is T₄ with coefficient 20.

How is the formula obtained? : In the expansion the first term is aⁿ (r = 0), the second aⁿ⁻¹b (r = 1) and so on; so ⁿCᵣ aⁿ⁻ʳ bʳ is the (r + 1)-th term, not the r-th.

  • Easy : The 5th term of (x + y)⁷: r = 4, T₅ = ⁷C₄ x³ y⁴ = 35x³y⁴.
  • Medium : The coefficient of x³ in (x + 2)⁶: r = 3, ⁶C₃ × 2³ = 20 × 8 = 160.
  • Exam-type : The term independent of x in (x² + 1/x)⁹: power of x is 18 − 3r = 0, r = 6, ⁹C₆ = 84.
Remember

For even n there is one middle term; for odd n there are two. The middle term of (1 + x)⁸ is 70x⁴.

Common mistake

Taking ⁿCᵣ … as the r-th term. It is the (r + 1)-th term.

Shortcut

For the term independent of x set up the power equation and check that r is a whole number. Memory clue : "Power zero, r found".

CET trap

If r is not a whole number or exceeds n, there is no such term.

CET-style example : The term independent of x in (x + 1/x)¹⁰ is? (A) 252 (B) 210 (C) 120 (D) 126

Answer : (A). The power of x is 10 − 2r = 0, r = 5; ¹⁰C₅ = 252. Time-saving method : With even power 10, r = n/2; take ¹⁰C₅ directly.

Spot the difference

nMiddle termsExample
evenone: T(n/2 + 1)(1 + x)⁸: T₅
oddtwo: T((n + 1)/2), T((n + 3)/2)(1 + x)⁷: T₄, T₅

Practice questions

Easy
  1. 1Write the 3rd term of (x + y)⁵.
  2. 2Write the middle term of (1 + x)⁸.
  3. 3What is the first term of (a + b)⁸?
Medium
  1. 4Find the coefficient of x⁴ in (x + 3)⁶.
  2. 5What is the coefficient of the middle term T₅ of (1 + x)⁸?
  3. 6Find the coefficients of the two middle terms of (1 + x)⁹.
Challenging
  1. 7Find the coefficient of x⁴ in (2x + 1)⁸.
  2. 8Find the coefficient of x⁴ in (x + 1/x)¹².
Show answers (try them yourself first)
  1. 1⁵C₂ x³y² = 10x³y².
  2. 270x⁴.
  3. 3a⁸.
  4. 4⁶C₂ × 3² = 15 × 9 = 135.
  5. 570.
  6. 6⁹C₄ = ⁹C₅ = 126.
  7. 7⁸C₄ × 2⁴ = 70 × 16 = 1120.
  8. 8x चा घात 12 − 2r = 4, r = 4 ; ¹²C₄ = 495.

Expansion for negative and fractional index

From daily life

1/(1 − x) looks like a fraction but, for small x, behaves like the endless sum 1 + x + x² + …. It helps to find roots and reciprocals without a calculator. Building on classes 9–10 : Since classes 9–10 we know negative and fractional indices; here we use them in the binomial formula.

Simple definition

If n is negative or a fraction and ∣x∣ < 1, then (1 + x)ⁿ = 1 + nx + n(n − 1)/2! x² + n(n − 1)(n − 2)/3! x³ + …; the terms never end.

Rule 1: Useful expansions — (1 − x)⁻¹ = 1 + x + x² + x³ + …; (1 + x)⁻¹ = 1 − x + x² − …; (1 − x)⁻² = 1 + 2x + 3x² + 4x³ + …; (1 + x)¹ᐟ² = 1 + x/2 − x²/8 + x³/16 − ….

11 term1.52 terms1.753 terms1.8754 terms1.93755 terms2(1−x)⁻¹ = 1 + x + x² + … for ∣x∣ < 1x = 0.5

In the figure : For x = 0.5 the sums 1, 1.5, 1.75, 1.875, … approach 2.

How is the formula obtained? : For negative n the product n(n − 1)… never becomes zero, so the terms do not end. For n = −1: (−1)(−2)/2! = 1, (−1)(−2)(−3)/3! = −1; hence (1 + x)⁻¹ = 1 − x + x² − ….

  • Easy : (1 + x)⁻¹ = 1 − x + x² − x³ + … (∣x∣ < 1).
  • Medium : (1 − 2x)⁻²: put y = 2x in (1 − y)⁻² = 1 + 2y + 3y² + 4y³: 1 + 4x + 12x² + 32x³ + …
  • Exam-type : (1.03)¹ᐟ³: x = 0.03, n = 1/3; 1 + 0.01 − 0.0001 + … ≈ 1.0099 (four decimals).
Remember

The formula holds only for ∣x∣ < 1; for (1 + 2x)ⁿ it needs ∣x∣ < 1/2.

Common mistake

Assuming the expansion has a last term. The terms never end; take only as many as needed.

Shortcut

Write any number as (1 + small): 1.03, 0.99 = 1 − 0.01. The first 3 terms usually suffice. Memory clue : "One plus small".

CET trap

In (1 − x)ⁿ the signs of the x terms flip; in (1 − x)⁻¹ all terms are positive.

CET-style example : The coefficient of x² in (1 − x)⁻³ is? (A) 6 (B) 3 (C) 9 (D) 12

Answer : (A). (−3)(−4)/2 × (−1)² = 6. Time-saving method : Put n = −3 in n(n − 1)/2; two negatives give +12/2.

Spot the difference

Pointn a positive integern negative / fraction
Termsn + 1 (they end)endless
Conditionany x∣x∣ < 1
Example(1 + x)⁴: 5 terms(1 + x)⁻¹: endless terms

Practice questions

Easy
  1. 1Write the first three terms of (1 − x)⁻¹.
  2. 2What is the coefficient of x in (1 + x)¹ᐟ²?
  3. 3For which x is the expansion of (1 + 2x)⁻¹ valid?
Medium
  1. 4Find the coefficient of x² in (1 + x)⁻².
  2. 5Write the first three terms of (1 − x)⁻¹ᐟ².
  3. 6Find 1/1.01 correct to four decimals.
Challenging
  1. 7Find the coefficient of x³ in (1 − 2x)⁻².
  2. 8Find √1.04 correct to four decimals.
Show answers (try them yourself first)
  1. 11 + x + x².
  2. 21/2.
  3. 3∣x∣ < 1/2.
  4. 4(−2)(−3)/2 = 3.
  5. 51 + x/2 + 3x²/8.
  6. 61 − 0.01 + 0.0001 − … ≈ 0.9901.
  7. 74 × 2³ = 32.
  8. 81 + 0.02 − 0.0002 + … ≈ 1.0198.

Lesson 4: Induction and binomial theorem — quick revision

TopicFormula / point
Inductionfoundation n = 1; assume n = k; prove n = k + 1; conclude all n
Sums1 + … + n = n(n + 1)/2; 1² + … + n² = n(n + 1)(2n + 1)/6; sum of odds n²
Divisibilityaⁿ − bⁿ by (a − b); 7ⁿ − 1 by 6; n³ − n by 6
Binomial theorem(a + b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ bʳ; n + 1 terms; coefficients add to 2ⁿ
General termTᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ; middle: even n one, odd n two
Negative / fractional(1 + x)ⁿ = 1 + nx + n(n − 1)/2! x² + …; ∣x∣ < 1

🎯 Most important points for the exam

  • Use the assumption in the succession step.
  • Tᵣ₊₁ is the (r + 1)-th term, not the r-th.
  • For the sum of coefficients put x = 1.
  • For negative or fractional index check ∣x∣ < 1.
  • For the term independent of x set the power of x to zero.
Study mantra

"Stand on k, jump to k + 1".