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If T 2 / T 3 in the Expansion of ( a + B ) N and T 3 / T 4 in the Expansion of ( a + B ) N + 3 Are Equal, Then N = (A) 3 (B) 4 (C) 5 (D) 6 - Mathematics

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प्रश्न

If  \[T_2 / T_3\]  in the expansion of \[\left( a + b \right)^n \text{ and } T_3 / T_4\]  in the expansion of \[\left( a + b \right)^{n + 3}\]  are equal, then n =

 
 

विकल्प

  • 3

  •  4

  •  5

  •  6

     
MCQ
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उत्तर

 5

\[\text{ In the expansion}  (a + b )^n , \text{ we have } \]

\[\frac{T_2}{T_3} = \frac{^{n}{}{C}_1 a^{n - 1} \times b^1}{^{n}{}{C}_2 a^{n - 2} \times b^2}\]

\[\text{ In the expansion } (a + b )^{n + 3} , \text{ we have } \]

\[\frac{T_3}{T_4} = \frac{^{n + 3}{}{C}_2 a^{n + 1} b^2}{^{n + 3}{}{C}_3 a^n b^3}\]

\[\text{ Thus, we have } \]

\[\frac{T_2}{T_3} = \frac{T_3}{T_4}\]

\[ \Rightarrow \frac{^{n}{}{C}_1 a}{^{n}{}{C}_2 b} = \frac{^{n + 3}{}{C}_2 a}{^{n + 3}{}{C}_3 b}\]

\[ \Rightarrow \frac{2}{n - 1} = \frac{3}{n + 1}\]

\[ \Rightarrow 2n + 2 = 3n - 3\]

\[ \Rightarrow n = 5\]

 
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Introduction of Binomial Theorem
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 18: Binomial Theorem - Exercise 18.4 [पृष्ठ ४७]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 18 Binomial Theorem
Exercise 18.4 | Q 19 | पृष्ठ ४७

संबंधित प्रश्न

Using binomial theorem, write down the expansions  :

(iii)  \[\left( x - \frac{1}{x} \right)^6\]

\[= ^{5}{}{C}_0 (2x )^5 (3y )^0 +^{5}{}{C}_1 (2x )^4 (3y )^1 + ^{5}{}{C}_2 (2x )^3 (3y )^2 + ^{5}{}{C}_3 (2x )^2 (3y )^3 + ^{5}{}{C}_4 (2x )^1 (3y )^4 +^{5}{}{C}_5 (2x )^0 (3y )^5\]

\[= 32 x^5 + 5 \times 16 x^4 \times 3y + 10 \times 8 x^3 \times 9 y^2 + 10 \times 4 x^2 \times 27 y^3 + 5 \times 2x \times 81 y^4 + 243 y^5 \]
\[ = 32 x^5 + 240 x^4 y + 720 x^3 y^2 + 1080 x^2 y^3 + 810x y^4 + 243 y^5 \]

 

 


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