Find the value of this sum (Canadian Math Olympiad, 1976)?

Question

Let a_0=1, a_1=2 and

for any n>=1, n*(n+1)*a_(n+1)=

=n*(n-1)*a_n - (n-2)*a_(n-1).

Calculate the value of

(a_0/a_1)+(a_1/a_2)+..+(a_50/a_51).

Here each underscore "_" sign means a subscript.

Only mathematically justified answers are welcome. Thank you.

Answer 1

i have an answer but i don't think it is mathematically justified..for the fact that i used indirect prooving

Answer 2

If you apply the recurrence relation a few times you get

a_2 = 1/2, a_3 = 1/6, a_4 = 1/24, a_5 = 1/120

So it looks as though

a_n = 1/n! for n > 1

I'm sure that this could be proven using induction.

Then a_n/a_(n+1) = n + 1

This means that the required sum is

1/2 + 4 + (3 + 4 + 5 .... + 51) = 4.5 + 51*52/2 - 3 = 1327.5

Answer 3

a[0]=1

a[1]=2

Rearranging the relation gives: a[n+1] = (n-1)/(n+1) a[n] - (n-2)/(n(n+1)) a[n-1]

Hence

a[2] = 0/2 * 2 - -1/(1*2) = 1/2

a[3] = 1/3 * 1/2 - 0/(2*3) * 2 = 1/6 - 0 = 1/6

a[4] = 2/4 * 1/6 - 1/(3*4) * 1/2 = 1/12 - 1/24 = 1/24

a[5] = 3/5 * 1/24 - 2/(4*5) * 1/6 = 1/40 - 1/60 = 1/120

It appears that a[n] = a[n-1] / n

To prove this mathematically lets use induction.

Initial case

(n=3)

a[3] = 1/6 = a[1]/3

So it is it true for n=3.

Assume it is true for and n=k

So a[k] = a[k-1] / k

Now

a[k+1] = (k-1)/(k+1) * a[k] - (k-2)/(k(k+1) * a[k-1] from the definition

= (k-1)/(k+1) * a[k] - (k-2)/(k(k+1)) * a[k]*k ... using the assumption

= ( (k-1)/(k+1) - (k-2)/(k+1)) * a[k]

= ((k-1 - (k-2)) / (k+1) * a[k]

= 1/(k+1) * a[k]

So if it is true for n=k then it is true for n=k+1, but it is true for n=3 so by the principle of mathematical induction it is true for all n>=3.

Sum(i=1 to 51) a[i-1]/a[i]

= a[0]/a[1] + a[1]/a[2] + Sum(i=3 to 51) a[i-1]/a[i]

= 1/2 + 4 + Sum(i=3 to 51) i

= 1/2 + 4 + 51*52/2 - 2*3/2

= 1327.5

Answer 4

What do the * mean? Are they multiplication?

This is definitely to do with factorials (work out the first few terms) You get T_n = n!/(n-1)! which is equal to n(n-1)! / (n-1)!

=n! You want to find the sum of 51 terms...

Sorry, that should be T_n = n, so you are looking for 1/2 + 4 +

(3+4+5+...+49)