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 Front to back (Posted on 2007-03-18)
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.

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 Solution Comment 3 of 3 |

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
 Posted by K Sengupta on 2007-03-19 11:34:16

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