If 2 n C 2-2 n C 4-2 n C 6-2 n C 2 n -511- then absolue value of n C 0- n C 1 31- n C 2 32-1 n n c n 3 n is

We know that nbsp; ^2nC_0+^2nC_2+^2nC_4+ ^2nC_6+⋯+^2 n C_2 n=2^2n 1 ⇒ 512=2^2n 1 ⇒ n=5 Let S=^n C_0 ^n C_1· 3^1+^n C_2·3^2+⋯+( 1)^n ^n C_n·3^n =∑_r=0^n ( 1)^ ...

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Byju's Answer

Standard XII

Mathematics

Sum of Binomial Coefficients with Alternate Signs

If 2 n C 2+2 ...

Question

If $^{2 n} C_{2} +^{2 n} C_{4} +^{2 n} C_{6} + \dots +^{2 n} C_{2 n} = 511$ , then absolue value of $^{n} C_{0} -^{n} C_{1} 3^{1} +^{n} C_{2} 3^{2} + \dots + (- 1)^{n}^{n} c_{n} 3^{n}$ is

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Solution

We know that $^{2 n} C_{0} +^{2 n} C_{2} +^{2 n} C_{4} +^{2 n} C_{6} + \dots +^{2 n} C_{2 n} = 2^{2 n - 1}$
$\Rightarrow 512 = 2^{2 n - 1}$
$\Rightarrow n = 5$

Let $S =^{n} C_{0} -^{n} C_{1} 3^{1} +^{n} C_{2} 3^{2} + \dots + (- 1)^{n}^{n} C_{n} 3^{n}$
$= n \sum r = 0 (- 1)^{r} \cdot^{n} C_{r} \cdot 3^{r}$
$= n \sum r = 0^{n} C_{r} (- 3)^{r} = (1 - 3)^{n} = (- 2)^{n}$
Since, $n = 5$
$∴ S = (- 2)^{5} = - 32$
$\Rightarrow | S | = 32$

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If 2nC2+2nC4+2nC6+⋯+ 2nC2n=511, then absolue value of nC0−nC131+nC232+⋯+(−1)n ncn3n is

We know that 2nC0+2nC2+2nC4+2nC6+⋯+2nC2n=22n−1 ⇒512=22n−1 ⇒n=5 Let S=nC0−nC131+nC232+⋯+(−1)n nCn3n =n∑r=0(−1)r⋅ nCr⋅3r =n∑r=0nCr(−3)r =(1−3)n=(−2)n Since, n=5 ∴S=(−2)5=−32 ⇒|S|=32

If $^{2 n} C_{2} +^{2 n} C_{4} +^{2 n} C_{6} + \dots +^{2 n} C_{2 n} = 511$ , then absolue value of $^{n} C_{0} -^{n} C_{1} 3^{1} +^{n} C_{2} 3^{2} + \dots + (- 1)^{n}^{n} c_{n} 3^{n}$ is