How would you solve: 3(2n+4)!/(2n+1)! = 2(n+4)!/n! Solve for n.

Question

Thanks, having a lot of trouble with this question.

Answer 1

3(2n+4)!/(2n+1)!=3(2n+4)(2n+3)(2n+2)(2n+1)!/(2n+1)!=3(2n+4)(2n+3)(2n+2);

2(n+4)!/n!=2(n+4)(n+3)(n+2)(n+1).

Therefore:

3(2n+4)(2n+3)(2n+2)=2(n+4)(n+3)(n+2)(n+1),

3[2(n+2)](2n+3)[2(n+1)]=2(n+4)(n+3)(n+2)(n+1),

12(2n+3)=2(n²+7n+12),

6(2n+3)=n²+7n+12,

12n+18=n²+7n+12,

n²-5n-6=(n+1)(n-6)=0 so n=-1 or 6. Consider n=6 only then check:

3(16!)/13!=3×16×15×14; 2(10!)/6!=2×10×9×8×7.

3×2×15×2=2×10×9, 180=180✔️