binomial theorem

Question

for all natural numbers n, {(2²ⁿ⁾*(3²ⁿ)}-1-35n is divisible by???

 

Answer 1

2²ⁿ*3²ⁿ-1-35n is divisible by what number(s)

2²ⁿ*3²ⁿ=(2*3)²ⁿ=6²ⁿ=(6²)ⁿ

6²=36 and 36-1=35 which is divisible by 5 and by 7 as well as itself, 35

36 raised to any power and then minus 1 gives a number divisible by 35

(36)ⁿ-1 is divisible by 35, or in other words, is a multiple of 35 for n=1,2,3,4...

if two numbers are both divisible by 35, then their sum or difference is divisible by 35

35n is a multiple of 35 for n=1,2,3,4,...

[(36)ⁿ-1] is a multiple of 35 and 35n is a multiple of 35, so their difference is a multiple of 35

[(36)ⁿ-1]-35n is a multiple of 35 so this difference is divisible by 35

also, for n=1,2,3, and 4 you get the following numbers

n=1, 36-1-35=0

n=2, 1296-1-70=1225

n=3, 46,656-1-105=46,550

n=4, 1,679,616-1-140=1,679,475

0, 1225, 46,550, and 1,679,475 are all divisible by 1225 as well as being divisible by 35

 

 

 

Answer 2

Minakshi:

 

I am adding something at the end.  Please read.

 

Write

 

2²ⁿ3²ⁿ = (6)²ⁿ = (36)ⁿ = (1 + 35)ⁿ.

 

(1 + 35)ⁿ = 1 + 35n + ∑_r=2^{n  n}C_r(35)^r

 

(1 + 35)ⁿ - 1 - 35 = (35²) [∑_r=2ⁿ (35)^{r - 2}].

 

The above expression shows that 35² = 1225 is a factor of the left side, that is [(2²ⁿ)(3²ⁿ)] - 1 - 35 is divisible by 1225.  Final answer.  (Nice problem!)

 

Dr. G.

 

 

P.S.  If you choose any integer k > 1, write kⁿ = [1 - (k-1)]ⁿ, write out the first two terms of the expansion and leave the rest as sum, subtract the first two terms from both sides, you get

 

kⁿ - 1 - n(k-1) = ∑_{r = 2} ^{n n}C_r (k-1)^r.

 

Now you write the term on the right s (k-1)² ∑_r=2  ⁿC_r  (k-1)^r-2.

 

This shows that the left side is divisible by (k-1)², for any K >1.

 

You had in the problem, k = 2² 3² = 36, so the answer was (36-1)² = 1225.