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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