Menu Close

please-help-me-to-solve-this-in-N-A-n-2-3C-n-n-2-n-20-

Tinku Tara 0 Comments
Pin




Question Number 85322 by mathocean1 last updated on 20/Mar/20
please help me to solve this in N A_n ^2 −3C_n ^( n−2) +n=−20
$${please}\:{help}\:{me}\:{to}\:{solve}\:{this}\:{in}\:\mathbb{N} \\ $$$${A}_{{n}} ^{\mathrm{2}} −\mathrm{3}{C}_{{n}} ^{\:{n}−\mathrm{2}} +{n}=−\mathrm{20} \\ $$
Commented by mr W last updated on 20/Mar/20
what do you want to get exactly? do you want to get A_n ? this is possible: A_n =±(√(3C_n ^( n−2) −n−20)) or do you want to solve for n? this is not possible, because it is not given what A_n is.
$${what}\:{do}\:{you}\:{want}\:{to}\:{get}\:{exactly}? \\ $$$$ \\ $$$${do}\:{you}\:{want}\:{to}\:{get}\:{A}_{{n}} ?\:{this}\:{is} \\ $$$${possible}:\:{A}_{{n}} =\pm\sqrt{\mathrm{3}{C}_{{n}} ^{\:{n}−\mathrm{2}} −{n}−\mathrm{20}} \\ $$$$ \\ $$$${or}\:{do}\:{you}\:{want}\:{to}\:{solve}\:{for}\:{n}?\:{this} \\ $$$${is}\:{not}\:{possible},\:{because}\:{it}\:{is}\:{not}\: \\ $$$${given}\:{what}\:{A}_{{n}} \:{is}. \\ $$
Commented by mathocean1 last updated on 20/Mar/20
A_n ^2 =((n!)/((n−2)!)) this is the formula that i have.
$${A}_{{n}} ^{\mathrm{2}} =\frac{\mathrm{n}!}{\left({n}−\mathrm{2}\right)!}\:\:{this}\:{is}\:{the}\:{formula}\:{that}\: \\ $$$${i}\:{have}. \\ $$
Commented by mr W last updated on 20/Mar/20
what do you mean with C_n ^( n−2) ?
$${what}\:{do}\:{you}\:{mean}\:{with}\:{C}_{{n}} ^{\:{n}−\mathrm{2}} ? \\ $$
Commented by mr W last updated on 20/Mar/20
for combination calculation C_r ^n is defined only for 0≤r≤n therefore C_n ^(n−2) is not possible! or do you mean (C_n )^(n−2) ? then you must specify what is C_n .
$${for}\:{combination}\:{calculation} \\ $$$${C}_{{r}} ^{{n}} \:{is}\:{defined}\:{only}\:{for}\:\mathrm{0}\leqslant{r}\leqslant{n} \\ $$$${therefore}\:{C}_{{n}} ^{{n}−\mathrm{2}} \:{is}\:{not}\:{possible}! \\ $$$${or}\:{do}\:{you}\:{mean}\:\left({C}_{{n}} \right)^{{n}−\mathrm{2}} ?\:{then}\:{you} \\ $$$${must}\:{specify}\:{what}\:{is}\:{C}_{{n}} . \\ $$
Commented by mr W last updated on 20/Mar/20
i think your question is not correct. please recheck!
$${i}\:{think}\:{your}\:{question}\:{is}\:{not}\:{correct}. \\ $$$${please}\:{recheck}! \\ $$
Commented by mathocean1 last updated on 20/Mar/20
I rechecked it sir...
$${I}\:{rechecked}\:{it}\:{sir}… \\ $$
Commented by mathocean1 last updated on 20/Mar/20
It is like this. I think that C_n ^(n−2) =((n!)/(2×(n−2)!))
$${It}\:{is}\:{like}\:{this}. \\ $$$${I}\:{think}\:{that}\:{C}_{{n}} ^{{n}−\mathrm{2}} =\frac{{n}!}{\mathrm{2}×\left({n}−\mathrm{2}\right)!} \\ $$
Commented by mr W last updated on 20/Mar/20
you write C_n ^(n−2) , but you mean C_(n−2) ^n .
$${you}\:{write}\:{C}_{{n}} ^{{n}−\mathrm{2}} ,\:{but}\:{you}\:{mean}\:{C}_{{n}−\mathrm{2}} ^{{n}} . \\ $$
Commented by mr W last updated on 20/Mar/20
so your question is to solve n for A_n ^2 −3C_(n−2) ^n +n=−20 with A_n ^2 =((n!)/((n−2)!)) ((n!)/((n−2)!))−3×((n!)/(2!(n−2)!))+n=−20 ((n!)/((n−2)!))−n=20 n(n−1)−n=20 n^2 −2n−20=0 n=1+(√(21)) ∉N ⇒no solution! again: the question is wrong!
$${so}\:{your}\:{question}\:{is}\:{to}\:{solve}\:{n}\:{for} \\ $$$${A}_{{n}} ^{\mathrm{2}} −\mathrm{3}{C}_{{n}−\mathrm{2}} ^{{n}} +{n}=−\mathrm{20} \\ $$$${with}\:{A}_{{n}} ^{\mathrm{2}} =\frac{{n}!}{\left({n}−\mathrm{2}\right)!} \\ $$$$ \\ $$$$\frac{{n}!}{\left({n}−\mathrm{2}\right)!}−\mathrm{3}×\frac{{n}!}{\mathrm{2}!\left({n}−\mathrm{2}\right)!}+{n}=−\mathrm{20} \\ $$$$\frac{{n}!}{\left({n}−\mathrm{2}\right)!}−{n}=\mathrm{20} \\ $$$${n}\left({n}−\mathrm{1}\right)−{n}=\mathrm{20} \\ $$$${n}^{\mathrm{2}} −\mathrm{2}{n}−\mathrm{20}=\mathrm{0} \\ $$$${n}=\mathrm{1}+\sqrt{\mathrm{21}}\:\notin\mathbb{N} \\ $$$$\Rightarrow{no}\:{solution}! \\ $$$$ \\ $$$${again}:\:{the}\:{question}\:{is}\:{wrong}! \\ $$
Commented by mathocean1 last updated on 20/Mar/20
yes
$${yes} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *