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3 years ago

Find a closed form for

1 answer:

5 0

$S_n=1\cdot1!+2\cdot2!+\cdots+n\cdot n!$
$S_n=(1+1-1)\cdot1!+(2+1-1)\cdot2!+\cdots+(n+1-1)\cdot n!$
$S_n=(2\cdot1!-1\cdot1!)+(3\cdot2!-1\cdot2!)+\cdots+\bigg((n+1)n!-\cdot n!\bigg)$
$S_n=(2!-1!)+(3!-2!)+\cdots+\bigg((n+1)!-1\cdot n!\bigg)$
$S_n=-1!+(n+1)!$
$S_n=(n+1)!-1$

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How much of a 50% Alcohol solution should we mix with 10 gallons of 35% solution to get 40% solution

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<h3>5 gallons of 50 % alcohol solution is used</h3>

<em><u>Solution:</u></em>

Let "x" is the gallons of 50 % alcohol solution

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Then, according to question,

"x" gallons of 50 % solution is mixed with 10 gallons of 35 % solution to get 10 + x gallons of 40 % solution

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