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Find the sum of the product of every pair of the first `n` of natural number |
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Answer» We find that `S=1*2+1*3+1*4+"...."+2*3+2*4+"......."+3*4+3*5+"....."(n-1)*n"......(i)"` `:.[1+2+3+"...."+(n-1)+n]^(2)=1^(2)+2^(2)+3^(2)+"...."+(n-1)^(2)+n^(2) +2[1*2+1*3+1*4+"...."+2*3+2*4+"...."+3*4+3*5+"......."+(n-1)*n]` `(sumn)^(2)=sumn^(2)+25" " [" from Eq. (i) "]` ` implies S=((sumn)^(2)=sumn^(2))/(2)` `({(n(n+1))/(2)}^(2)-(n(n+1)(2n+1))/(6))/(2)` `((n^(2)(n+1)^(2))/(4)-(n(n+1)(2n+1))/(6))/(2)` ,`(n(n+1))/(24)[3n(n+1)-2(2n+1)]` `(n(n+1)(3n^(2)-n+2))/(24)` Hence, `S=((n-1)n(n+1)(3n+2))/(24)` |
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