Question

show that (5n!)/((40^n)n!) is a natural number where n is a natural number.​

Answer 1

Step-by-step explanation:

n mathematics, the natural numbers are those used for counting (as in "there are six coins on the table") and ordering (as in "this is the third largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers".

Answer 2

Answer:

For prime p and positive integer n , the highest power νp(n!) is given by

νp(n!)=∑k=1∞⌊npk⌋.…(⋆)

Therefore, if p≠2,5 , the highest power of p dividing (5n)!/40n⋅n! equals

νp((5n)!)−νp(n!)=∑k=1∞(⌊5npk⌋−⌊npk⌋)≥0.

The highest power of 2 dividing (5n)!/40n⋅n! equals

ν2((5n)!)−ν2(n!)−ν2(40n)=∑k=1∞(⌊5n2k⌋−⌊n2k⌋)−ν2(8n)≥(⌊5n2⌋−⌊n2⌋)+(⌊5n4⌋−⌊n4⌋)−3n=0.

The highest power of 5 dividing (5n)!/40n⋅n! equals

ν5((5n)!)−ν5(n!)−ν5(40n)=∑k=1∞⌊5n5k⌋−∑k=1∞⌊n5k⌋−ν5(5n)=n+∑k=2∞⌊5n5k⌋−∑k=1∞⌊n5k⌋−n=0.

Thus, the highest power of p dividing (5n)!/(40n⋅n!) is ≥0 for each prime p , thereby proving (5n)!/(40n⋅n!) is a positive integer. ■