Choice (a) is correct. We know that the coefficient of xʳ in the expansion of (1 + x)ⁿ is ⁿCᵣ The coefficient of xᵖ in the expansion of (1 + x) ᵖ ⁺ ᑫ is ᵖ ⁺ ᑫCₚ and the coefficient of xᑫ in the expansion of (1 + x)ᵖ ⁺ ᑫ is ᵖ ⁺ ᑫCₚ = ᵖ ⁺ ᑫCₚ Hence, the coefficient of xᵖ and xᑫ in the expansion of (1Read more
Choice (a) is correct.
We know that the coefficient of xʳ in the expansion of (1 + x)ⁿ is ⁿCᵣ
The coefficient of xᵖ in the expansion of (1 + x) ᵖ ⁺ ᑫ is ᵖ ⁺ ᑫCₚ
and the coefficient of xᑫ in the expansion of (1 + x)ᵖ ⁺ ᑫ is ᵖ ⁺ ᑫCₚ = ᵖ ⁺ ᑫCₚ
Hence, the coefficient of xᵖ and xᑫ in the expansion of (1 + x)ᵖ ⁺ ᑫ are equal.
This question related to Chapter 7 maths Class 11th NCERT. From the Chapter 7: Binomial Theorem. Give answer according to your understanding.
Choice (b) is correct. We know that (r + 1) term in the binomial expansion of (1 + x)ⁿ is given by Tᵣ ₊ ₁ = ⁿCᵣ(1)ⁿ ⁻ ʳ xʳ = ⁿCᵣxʳ In the binomial expansion of (1 + x)¹⁰⁰ is given by Tᵣ ₊ ₁ = ¹⁰⁰Cᵣxʳ For coefficient of x⁵⁰, put r = 50 in above, we have T ₅₀ ₊ ₁ = ¹⁰⁰ C₅₀ x⁵⁰ The coefficient x⁵⁰ = Read more
Choice (b) is correct.
We know that (r + 1) term in the binomial expansion of (1 + x)ⁿ is given by
Tᵣ ₊ ₁ = ⁿCᵣ(1)ⁿ ⁻ ʳ xʳ = ⁿCᵣxʳ
In the binomial expansion of (1 + x)¹⁰⁰ is given by
Tᵣ ₊ ₁ = ¹⁰⁰Cᵣxʳ
For coefficient of x⁵⁰, put r = 50 in above, we have T ₅₀ ₊ ₁ = ¹⁰⁰ C₅₀ x⁵⁰
The coefficient x⁵⁰ = ¹⁰⁰ C₅₀
This question related to Chapter 7 maths Class 11th NCERT. From the Chapter 7: Binomial Theorem. Give answer according to your understanding.
Choice (a) is correct. We have (a + b + c)ⁿ = {a + (b + c)}ⁿ = aⁿ + ⁿC₁aⁿ ⁻ ¹ (b + c) + ⁿC₂aⁿ ⁻ ²(b - c)² + .... + ⁿCₙ(b + c)ⁿ On expanding each term of R.H.S., we get Number of terms in first term = 1 Number of terms in second term = 2 Number of terms in third term = 3 Number of terms in fourth teRead more
Choice (a) is correct.
We have (a + b + c)ⁿ = {a + (b + c)}ⁿ = aⁿ + ⁿC₁aⁿ ⁻ ¹ (b + c) + ⁿC₂aⁿ ⁻ ²(b – c)² + …. + ⁿCₙ(b + c)ⁿ
On expanding each term of R.H.S., we get
Number of terms in first term = 1
Number of terms in second term = 2
Number of terms in third term = 3
Number of terms in fourth term = 4
Number of terms in (n + 1)th term = n + 1
Total number of terms = 1 + 2 + 3 + ….. + (n +1)
= (n – 1) {1 + (n + 1)}/2
= (n + 1)(n + 2) /2
This question related to Chapter 7 maths Class 11th NCERT. From the Chapter 7: Binomial Theorem. Give answer according to your understanding.
Choice (c) is correct. We know that the total number of terms in the expansion of (x + a)ⁿ is (n + 1). Total number of terms in the expansion of (1 + x)¹⁰⁰¹ is (1001 + 1) = 1002. Total number of terms in the expansion of (1 + x) ¹⁰⁰¹ -x¹⁰⁰¹ is 1001 as two terms cancel each other This question relateRead more
Choice (c) is correct.
We know that the total number of terms in the expansion of (x + a)ⁿ is (n + 1).
Total number of terms in the expansion of (1 + x)¹⁰⁰¹ is (1001 + 1) = 1002.
Total number of terms in the expansion of (1 + x) ¹⁰⁰¹ -x¹⁰⁰¹ is 1001 as two terms cancel each other
This question related to Chapter 7 maths Class 11th NCERT. From the Chapter 7: Binomial Theorem. Give answer according to your understanding.
The coefficients of xᵖ and xᑫ, p and q are positive integers in the expansion of (1 + x)ᵖ ⁺ ᑫ are
Choice (a) is correct. We know that the coefficient of xʳ in the expansion of (1 + x)ⁿ is ⁿCᵣ The coefficient of xᵖ in the expansion of (1 + x) ᵖ ⁺ ᑫ is ᵖ ⁺ ᑫCₚ and the coefficient of xᑫ in the expansion of (1 + x)ᵖ ⁺ ᑫ is ᵖ ⁺ ᑫCₚ = ᵖ ⁺ ᑫCₚ Hence, the coefficient of xᵖ and xᑫ in the expansion of (1Read more
Choice (a) is correct.
We know that the coefficient of xʳ in the expansion of (1 + x)ⁿ is ⁿCᵣ
The coefficient of xᵖ in the expansion of (1 + x) ᵖ ⁺ ᑫ is ᵖ ⁺ ᑫCₚ
and the coefficient of xᑫ in the expansion of (1 + x)ᵖ ⁺ ᑫ is ᵖ ⁺ ᑫCₚ = ᵖ ⁺ ᑫCₚ
Hence, the coefficient of xᵖ and xᑫ in the expansion of (1 + x)ᵖ ⁺ ᑫ are equal.
This question related to Chapter 7 maths Class 11th NCERT. From the Chapter 7: Binomial Theorem. Give answer according to your understanding.
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The coefficient of x⁵⁰ in (1 + x)¹⁰⁰ is
Choice (b) is correct. We know that (r + 1) term in the binomial expansion of (1 + x)ⁿ is given by Tᵣ ₊ ₁ = ⁿCᵣ(1)ⁿ ⁻ ʳ xʳ = ⁿCᵣxʳ In the binomial expansion of (1 + x)¹⁰⁰ is given by Tᵣ ₊ ₁ = ¹⁰⁰Cᵣxʳ For coefficient of x⁵⁰, put r = 50 in above, we have T ₅₀ ₊ ₁ = ¹⁰⁰ C₅₀ x⁵⁰ The coefficient x⁵⁰ = Read more
Choice (b) is correct.
We know that (r + 1) term in the binomial expansion of (1 + x)ⁿ is given by
Tᵣ ₊ ₁ = ⁿCᵣ(1)ⁿ ⁻ ʳ xʳ = ⁿCᵣxʳ
In the binomial expansion of (1 + x)¹⁰⁰ is given by
Tᵣ ₊ ₁ = ¹⁰⁰Cᵣxʳ
For coefficient of x⁵⁰, put r = 50 in above, we have T ₅₀ ₊ ₁ = ¹⁰⁰ C₅₀ x⁵⁰
The coefficient x⁵⁰ = ¹⁰⁰ C₅₀
This question related to Chapter 7 maths Class 11th NCERT. From the Chapter 7: Binomial Theorem. Give answer according to your understanding.
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The number of terms in the expansion of (a + b + c)ⁿ, where n ∈ N is
Choice (a) is correct. We have (a + b + c)ⁿ = {a + (b + c)}ⁿ = aⁿ + ⁿC₁aⁿ ⁻ ¹ (b + c) + ⁿC₂aⁿ ⁻ ²(b - c)² + .... + ⁿCₙ(b + c)ⁿ On expanding each term of R.H.S., we get Number of terms in first term = 1 Number of terms in second term = 2 Number of terms in third term = 3 Number of terms in fourth teRead more
Choice (a) is correct.
We have (a + b + c)ⁿ = {a + (b + c)}ⁿ = aⁿ + ⁿC₁aⁿ ⁻ ¹ (b + c) + ⁿC₂aⁿ ⁻ ²(b – c)² + …. + ⁿCₙ(b + c)ⁿ
On expanding each term of R.H.S., we get
Number of terms in first term = 1
Number of terms in second term = 2
Number of terms in third term = 3
Number of terms in fourth term = 4
Number of terms in (n + 1)th term = n + 1
Total number of terms = 1 + 2 + 3 + ….. + (n +1)
= (n – 1) {1 + (n + 1)}/2
= (n + 1)(n + 2) /2
This question related to Chapter 7 maths Class 11th NCERT. From the Chapter 7: Binomial Theorem. Give answer according to your understanding.
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If (1 – x + x²)ⁿ = a₀ + a₁x + a₂x² + a₃x³ + …. + a₂ₙx²ⁿ, then a₁ + a₃ + a₅ …. +a₂ₙ ₋ ₁ equals
Choice (c) is correct. We have, (1 + x + x²)ⁿ = a₀ + a₁x + a₂x² + a₃x³ + .... + a₂ₙx²ⁿ Putting x = 1 and - 1 in (1), we get 1 = a₀ + a₁ + a₂ + a₃ + .... a₂ₙ and 3ⁿ = a₀ + a₁x + a₂x² + a₃x³ + .... + a₂ₙ Subtracting (3) from (2), we get 1 - 3ⁿ = 2(a₁ + a₃ + a₅ .... +a₂ₙ ₋ ₁) ⇒ a₁ + a₃ + a₅ .... +a₂ₙRead more
Choice (c) is correct.
We have, (1 + x + x²)ⁿ = a₀ + a₁x + a₂x² + a₃x³ + …. + a₂ₙx²ⁿ
Putting x = 1 and – 1 in (1), we get
1 = a₀ + a₁ + a₂ + a₃ + …. a₂ₙ
and 3ⁿ = a₀ + a₁x + a₂x² + a₃x³ + …. + a₂ₙ
Subtracting (3) from (2), we get
1 – 3ⁿ = 2(a₁ + a₃ + a₅ …. +a₂ₙ ₋ ₁)
⇒ a₁ + a₃ + a₅ …. +a₂ₙ ₋ ₁ = 1 – 3ⁿ/ 2
This question related to Chapter 7 maths Class 11th NCERT. From the Chapter 7: Binomial Theorem. Give answer according to your understanding.
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The total number of terms in the expansion of (1 + x)¹⁰⁰¹ -x¹⁰⁰¹ is :
Choice (c) is correct. We know that the total number of terms in the expansion of (x + a)ⁿ is (n + 1). Total number of terms in the expansion of (1 + x)¹⁰⁰¹ is (1001 + 1) = 1002. Total number of terms in the expansion of (1 + x) ¹⁰⁰¹ -x¹⁰⁰¹ is 1001 as two terms cancel each other This question relateRead more
Choice (c) is correct.
We know that the total number of terms in the expansion of (x + a)ⁿ is (n + 1).
Total number of terms in the expansion of (1 + x)¹⁰⁰¹ is (1001 + 1) = 1002.
Total number of terms in the expansion of (1 + x) ¹⁰⁰¹ -x¹⁰⁰¹ is 1001 as two terms cancel each other
This question related to Chapter 7 maths Class 11th NCERT. From the Chapter 7: Binomial Theorem. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-7