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If a and B Are Coefficients of Xn in the Expansions of ( 1 + X ) 2 N and ( 1 + X ) 2 N − 1 Respectively, Then Write the Relation Between a and B. - Mathematics

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प्रश्न

If a and b are coefficients of xn in the expansions of \[\left( 1 + x \right)^{2n} \text{ and } \left( 1 + x \right)^{2n - 1}\] respectively, then write the relation between a and b.

 
 

उत्तर

\[\text{ Coefficient of } x^n \text{ in the expansion}  (1 + x )^{2n} =^{2n}{}{C}_n = a\]

\[\text{ Coefficient of } x^n \text{ in the expansion}  (1 + x )^{2n - 1} = ^{2n - 1}{}{C}_n = b\]

\[\text{ Now, we have:}  \]

\[ ^{2n}{}{C}_n = \frac{2n!}{n! . n!} = \frac{2n(2n - 1)!}{n\left( n - 1 \right)! n!} . . . \left( 1 \right)\]

\[ \text{ and }  ^{2n - 1}{}{C}_n = \frac{(2n - 1)!}{n!(n - 1)!} . . . \left( 2 \right)\]

\[\text{ Dividing equation }  \left( 1 \right) \text{ by }  \left( 2 \right), \text{ we get } \]

\[ \Rightarrow \frac{^{2n}{}{C}_n}{^{2n - 1}{}{C}_n} = \frac{2n(2n - 1)! n! (n - 1)!}{n\left( n - 1 \right)! n! (2n - 1)!}\]

\[ \Rightarrow \frac{a}{b} = 2\]

\[ \Rightarrow a = 2b\]

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Introduction of Binomial Theorem
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 18: Binomial Theorem - Exercise 18.3 [पृष्ठ ४५]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 18 Binomial Theorem
Exercise 18.3 | Q 7 | पृष्ठ ४५

संबंधित प्रश्न

Using binomial theorem, write down the expansions  :

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

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

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