If -k-0434-k4-k- xkk- - 323- then the largest value of x is

∑_k=0^4(3^4 k(4 k)!) (x^kk! ) nbsp;=∑_k=0^4(3^4 k(4 k)!) (x^kk! )4!4! nbsp;= nbsp;∑_k=0^4(4!· 3^4 kx^k(4 k)!k!)14! nbsp;= nbsp;∑_k=0^4^4C_k· 3^4 k· ...

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Standard XII

Mathematics

Pascal Triangle

If ∑k=0434-k4...

Question

If $4 \sum k = 0 (\frac{3^{4 - k}}{(4 - k)!}) (\frac{x^{k}}{k!}) = \frac{32}{3},$ then the largest value of $x$ is

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Solution

$4 \sum k = 0 (\frac{3^{4 - k}}{(4 - k)!}) (\frac{x^{k}}{k!}) = 4 \sum k = 0 (\frac{3^{4 - k}}{(4 - k)!}) (\frac{x^{k}}{k!}) \frac{4!}{4!} = 4 \sum k = 0 (\frac{4! \cdot 3^{4 - k} x^{k}}{(4 - k)! k!}) \frac{1}{4!} = 4 \sum k = 0 \frac{^{4} C_{k} \cdot 3^{4 - k} \cdot x^{k}}{4!} = \frac{(3 + x)^{4}}{4!}$
Therefore,
$\frac{(3 + x)^{4}}{4!} = \frac{32}{3} \Rightarrow (3 + x)^{4} = 2^{8} \Rightarrow (x + 3) = \pm 4$
$\Rightarrow x = 1 or - 7$
So, the largest real value of $x$ is $1.$

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Pascal Triangle

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If 4∑k=0(34−k(4−k)!)(xkk!)=323, then the largest value of x is

4∑k=0(34−k(4−k)!)(xkk!)=4∑k=0(34−k(4−k)!)(xkk!)4!4!=4∑k=0(4!⋅34−kxk(4−k)!k!)14!=4∑k=04Ck⋅34−k⋅xk4!=(3+x)44! Therefore, (3+x)44!=323⇒(3+x)4=28⇒(x+3)=±4 ⇒x=1 or −7 So, the largest real value of x is 1.

If $4 \sum k = 0 (\frac{3^{4 - k}}{(4 - k)!}) (\frac{x^{k}}{k!}) = \frac{32}{3},$ then the largest value of $x$ is