Mathematical Odds and Ends

Sum of Product of All pairs of First n natural Numbers

We Know that 

(a+ b+c)² =  a² + b²+ c² +2(ab +bc +ca)

(a+ b+c)² =  a² + b²+ c² +2 ( Sum of product of all pairs of a,b and c)

Similarly we know that

(a+ b+c +d)² =  a² + b²+ c² + d+ 2 (ab +ac +ad +bc + bd +cd + da)

(a+ b+c +d)² =  a² + b²+ c² + d+ 2 (Sum of product of all pairs of a,b ,c and d, taken two at a time.)

Suppose now we have to find sum of product of all pairs of first n natural nos.; then we can write

(1+2+3+4+…..+n)²  =  (1²+2²+3²+4²+…..+n²) + 2(Sum of product of all pairs of first n natural numbers)

So now we can say (Sum of product of all pairs of first n natural numbers)

= { n(n+1)/2} – n(n+1)(2n+1)/6

=…..

=…..

= (n-1)n(n+1)(3n+2)/24

Now suppose we have to find sum of products of all pairs of first 10 natural Numbers all we have to do is to put n=10 in the above formula.

Thus we get the value as equal to (10-1)(10) (10+1) (3x10 +2)/24=9x10x11x32/24= 1320 
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