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Isotopes of an element have
(c) Isotopes of an element have different atomic masses. As the number of protons is the same, number of neutrons is different. https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-4/
(c) Isotopes of an element have different atomic masses. As the number of protons is the same, number of neutrons is different.
https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-4/
See lessIf 79⁴⁰ + 80⁴⁰ are compared then
Choice (b) is correct. 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: https://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-6
Choice (b) is correct.
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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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-6
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.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-7
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.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-7
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.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-7