Suppose one sequence is formed by multiplying termwise the sequence 1,2,...n by the sequence n,n-1,...1 and another sequence is formed by multiplying termwise the sequence 2,4,...n by the sequence 2,4,...n.

Show that for even n, the sum of each sequence is the same.

Since, n is even; n=2p (say)

Let the sum of the first sequence and the sum of the second sequence as described in the problem be respectively denoted by S and T.

Then,

S = 1*2p + 2*(2p-1) + ......+ (2p-1)*2 + 2p*1

= [1*2p + 2*(2p-1) + ......p(p+1)]+ [(p+1)*p+ ....+2p-1)*2 + 2p*1)

= 2[1*2p + 2*(2p-1) + ......p(p+1]

= 2*U, where:

U = [1*2p + 2*(2p-1) + ......p(p+1]

= Sum(i = 1 to p) i*(2p+1-i)

= (2p+1)Sum(i= 1 to p) i - Sum( i= 1 to p)i^2

= p(p+1)(2p+1)/2 - p(p+1)(2p+1)/6

= p(p+1)(2p+1)/3

Hence, S = 2p(p+1)(2p+1)/3

Again:

T = 4+ 16 + 36 + .....+ 4*p^2

= 4( 1+4+9+....+ p^2)

= 4p(p+1)(2p+1)/6

= 2p(p+1)(2p+1)/3

Therefore, S is always equal to T whenever n is even.

*Edited on ***March 19, 2007, 1:41 pm**