Question

Expand using Binomial Therom (x^2+3)^6​

Answer 1

EXPLANATION.

Expand binomial Theorem,

⇒ (x² + 3)⁶.

As we know that,

Binomial Theorem for positive integral index.

⇒ (x + a)ⁿ = ⁿC₀xⁿa⁰ + ⁿC₁xⁿ⁻¹a + ⁿC₂xⁿ⁻²a₂ + ===== + ⁿCₙxaⁿ.

Using this formula, we get.

⇒ (x² + 3)⁶ = ⁶C₀(x²)⁶(3)⁰ + ⁶C₁(x²)⁵(3)¹ + ⁶C₂(x²)⁴(3)² + ⁶C₃(x²)³(3)³ + ⁶C₄(x²)²(3)⁴ + ⁶C₅(x²)¹(3)⁵ + ⁶C₆(x²)⁰(3)⁶.

$\sf \implies ^{n}C_r = \dfrac{n!}{r! (n - r)!}$

Using this formula, we get.

⇒ ⁶C₀ = 6!/0! (6 - 0)! = 1.

⇒ ⁶C₁ = 6!/1! (6 - 1)! = 6!/5! = 6.

⇒ ⁶C₂ = 6!/2! (6 - 2)! = 15.

⇒ ⁶C₃ = 6!/3! (6 - 3)! = 20.

⇒ ⁶C₄ = 6!/4! (6 - 4)! = 15.

⇒ ⁶C₅ = 6!/5! (6 - 5)! = 6.

⇒ ⁶C₆ = 6!/6! (6 - 6)! = 1.

⇒ (x² + 3)⁶ = x¹² + 18x¹⁰ + 135x⁸ + 540x⁶ + 1215x⁴ + 1458x² + 729.

                                                                                                                                   

MORE INFORMATION.

Middle term in the expansion of (x + a)ⁿ.

(1) = If n is even then middle term = (n/2 + 1)th term.

(2) = If n is odd then middle terms are = (n + 1/2)th and (n + 3/2)th term binomial coefficient of middle term is the greatest binomial coefficient.

Answer 2

★$\huge\bf{\blue{\underline{Question:-}}}$

• Expand using Binomial Therom (x^2+3)^6

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★$\huge\bf{\red{\underline{Answer:-}}}$

☯$\large\bf{\red{(x^{2}+3)^{6}}}$

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★$\huge\bf{\purple{\underline{Explanation:-}}}$

☯$\large\sf{\orange{It's\: easy \:to \:generate \:the \:1\: 6...line \:of\: Pascal's\: Triangle,}} \\ \large\sf{\orange{which \:is \:the\: one\: we \:need. \:We \:have\: ones \:along \:the}}\\ \large\sf{\orange{edges \:and\: each \:number \:is \:the \:sum\: of\: the \:two \:above\: it.\:}}$

$\bf\boxed{\:\:\:\:\:\:\:1}$

$\bf\boxed{\:\:\:\:\:\:1\:\:1}$

$\bf\boxed{\:\:\:\:\:1\:\:2\:\:1}$

$\bf\boxed{\:\:\:\:1\:\:3\:\:3\:\:1}$

$\bf\boxed{\:\:\:1\:\:4\:\:6\:\:4\:\:1}$

$\bf\boxed{\:15\:\:10\:\:10\:\:5\:\:1}$

$\bf\boxed{1\:\:6\:\:15\:20\:15\:6\:1}$

☯$\large\sf{\orange{Now \:we \:can \:just\: write}}$

☯$\large\sf{(a+b)^{6}=a^{6}+6a^{5}b+15a^{4}b^{2}+20a^{3}b^{3}+15a^{2}b^{4}+6ab^{5}+b^{6}}$

☯$\large\sf{\orange{Note\: how\: the\: exponents\: on\: \bf{a}\: count\: down\: and \:the}} \\ \large\sf{\orange{ones \:on \: \bf{b}\: count \:up\: so \:every \:term\: is \:sixth \:degree.}}$

☯$\large\sf{\orange{In\: our\: problem\: we\: have\: a=x2\: and\: b=3 \:so\: we\:substitute:}}$

☯$\large\sf{(x^{2}+3)^{6}=(x^{2})^{6}+6(x^{2})^{5}(3)+15(x^{2})^{4}3^{2}+20(x^{2})^{3}3^{3}+15(x^{2})^{2}3^{4}+6(x^{2})3^{5}+3^{6}}$

☯$\large\sf{\orange{Some\: arithmetic\: later,}}$

➙${(x^{2}+3)^{6}=x^{12}+18x^{10}+135x^{8}+540x^{6}+1215x^{4}+1458x^{2}+729}$

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