If m=C2n, then C2m is equal to
3·C4n
C4n+1
3·C4n+1
3·C3n+1
3·C2n+1
The explanation for the correct option
It is given that m=C2n.
⇒m=n!2!×n-2!⇒m=nn-12
Thus, C2m=m!2!×m-2!
⇒C2m=mm-12⇒C2m=nn-12nn-12-12⇒C2m=nn-12n2-n-222⇒C2m=nn-12n2-2n+n-222⇒C2m=nn-12nn-2+1n-222⇒C2m=nn-12n-2n+122⇒C2m=n+1nn-1n-28⇒C2m=n+1nn-1n-2n-3!8n-3!⇒C2m=3n+1!3×8n+1-4!⇒C2m=3·n+1!24n+1-4!⇒C2m=3·n+1!4!×n+1-4!⇒C2m=3·C4n+1∵Cmn=n!m!n-m!
Thus, C2m is equal to 3·C4n+1.
Hence, (C) is the correct option.
If α= mC2, then find the value of αC2.