The largest value for x such that ∑k=0454−k(4−k)!(xkk!)=83 is

∑k=04(54−k(4−k)!)(xkk!)=∑k=04(54−k(4−k)!)(Xkk!)4!4! =∑k=04 4Ck.54−k.xk4!=(5+x)44! Now, (5+x)44!=83⇒(5+x)4=64 x+5=23/2=22 ⇒x=22−5

The largest value for x such that ∑k=0454−k(4−k)!(xkk!)=83 is

∑k=04(54−k(4−k)!)(xkk!)=∑k=04(54−k(4−k)!)(Xkk!)4!4! =∑k=04 4Ck.54−k.xk4!=(5+x)44! Now, (5+x)44!=83⇒(5+x)4=64 x+5=23/2=22 ⇒x=22−5

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The largest value for x such that $\sum_{k = 0}^{4} \frac{5^{4 - k}}{(4 - k)!} (\frac{x^{k}}{k!}) = \frac{8}{3}$ is

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$2 \sqrt{2} + 5$

$2 \sqrt{2} - 5$

$- 2 \sqrt{2} - 5$

$- 2 \sqrt{2} + 5$

answer is B.

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Detailed Solution

$\sum_{k = 0}^{4} (\frac{5^{4 - k}}{(4 - k)!}) (\frac{x^{k}}{k!}) = \sum_{k = 0}^{4} (\frac{5^{4 - k}}{(4 - k)!}) (\frac{X^{k}}{k!}) \frac{4!}{4!} = \sum_{k = 0}^{4} \frac{^{4} C_{k} {.5}^{4 - k} . x^{k}}{4!} = \frac{{(5 + x)}^{4}}{4!} N o w, \frac{{(5 + x)}^{4}}{4!} = \frac{8}{3} \Rightarrow {(5 + x)}^{4} = 64 x + 5 = 2^{3 / 2} = 2 \sqrt{2} \Rightarrow x = 2 \sqrt{2} - 5$