Question 90:
If 2n+2−2n+1+2n=C×2n, then find C.
We have,
2n+2−2n+1+2n=C×2n
2n+2−2n+1+2n=C×2n⇒2n.22−2n.21+2n=C×2n [∵am+n=am×an]⇒2n[22−21+1]=C×2n [taking common 2n in LHS]⇒2n[4−2+1]=C×2n⇒3×2n=C×2n 3×2n×2−n=C×2n×2−n [multiplying both sides by 2−n]⇒3×2n−n=C×2n−n [∵am×an=am+n]⇒3×20=C×20⇒3×1=C×1∴3=C [∵a0=1]