The sum of co-effcients in the expansion of (x1+2x)n is equal to 729. Then, the constant term in the expansion is :

(1+x)n= nC0nCPx+ nC2x2+.......nCnxnpart x=2(1+2)n= nC0+ nC1(2)+22nC2+.....2nCn....(1)(x1+2x)n=(x1)nnCo+nC1(x1)n−1(2x)+.....nCn(2x)nSum of coefficient = nCo+2nC1+......2nnCx 729=(1+2)n from (1)729=3n⇒36=3n⇒n=6(x1+2x)6=(x1)66C0+ 6C1(x1)5(2x)1+ 6C2(x1)4(2x)2+6C3(x1)3(2x)3+ 6C4(x1)2(2x)4+6C4(x1)2(2x)5+6C6(2x)6Constant term =6C3×23 =3!3!6!×2×2×2 ⇒6×6720×2×2×2 ⇒160The constant term value is 160

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The sum of co-effcients in the expansion of ({1}{x} + 2x )^n i

Question

The sum of co-effcients in the expansion of $(\frac{1}{x} + 2 x)^{n}$ is equal to $729$ . Then, the constant term in the expansion is :

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160

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The sum of co-effcients in the expansion of

(\frac{1}{x} + 2 x)^{n}

is equal to

729

. Then, the constant term in the expansion is :

8 mins ago

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Text solutionVerified

(1 + x)^{n} =^{n} C_{0}^{n} C_{P} x +^{n} C_{2} x^{2} + ...... .^{n} C_{n} x^{n}

part

x = 2

(1 + 2)^{n} =^{n} C_{0} +^{n} C_{1} (2) + 2^{2}^{n} C_{2} + .... .^{2^{n}} C_{n}

....(1)

(\frac{1}{x} + 2 x)^{n} = (\frac{1}{x})^{n}^{n} C_{o} +^{n} C_{1} (\frac{1}{x})^{n - 1} (2 x) + .... .^{n} C_{n} (2 x)^{n}

Sum of coefficient

=^{n} C_{o} + 2^{n} C_{1} + ...... 2^{n}^{n} C_{x}

729 = (1 + 2)^{n}

from (1)

729 = 3^{n}

\Rightarrow 3^{6} = 3^{n} \Rightarrow n = 6

(\frac{1}{x} + 2 x)^{6} = (\frac{1}{x})^{6}^{6} C_{0} +^{6} C_{1} (\frac{1}{x})^{5} (2 x)^{1} +^{6} C_{2} (\frac{1}{x})^{4} (2 x)^{2} +

^{6} C_{3} (\frac{1}{x})^{3} (2 x)^{3} +^{6} C_{4} (\frac{1}{x})^{2} (2 x)^{4} +^{6} C_{4} (\frac{1}{x})^{2} (2 x)^{5} +^{6} C_{6} (2 x)^{6}

Constant term

=^{6} C_{3} \times 2^{3}

= \frac{6 !}{3 ! 3 !} \times 2 \times 2 \times 2

\Rightarrow \frac{720}{6 \times 6} \times 2 \times 2 \times 2

\Rightarrow 160

The constant term value is

160

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Question Text	The sum of co-effcients in the expansion of $(\frac{1}{x} + 2 x)^{n}$ is equal to $729$ . Then, the constant term in the expansion is :
Answer Type	Text solution:1
Upvotes	150

The sum of co-effcients in the expansion of (x1​+2x)n is equal to 729. Then, the constant term in the expansion is :

Text solutionVerified

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The sum of co-effcients in the expansion of $(\frac{1}{x} + 2 x)^{n}$ is equal to $729$ . Then, the constant term in the expansion is :