Sum of products of first n natural numbers taken two at a time is
n(n2-1)(3n+2)24
n(n+1)2(3n+2)74
n2(n+1)(3n+2)48
n(n+1)(n+2)(3n+2)96
Explanation for the correct option:
Finding the sum of n natural numbers taken two at a time:
We know that,
x1+x2+x3+.......+xn=x12+x22+............+xn2+2x1x2+x2x3+.........+xn-1xnOnsubstitutingx1=1,x2=2,x3=3,........,xn=n,weget1+2+3+.......+n2=12+22+32+......n2+21.2+2.3+.........or1+2+3+.......+n2-12+22+......n2=2SWhenS=SumofproductofnumberstakentwoatatimeTherefore;2S=nn+122-nn+12n+16=n2n+124-nn+12n+16=nn+1nn+14-2n+16=nn+1×123nn+1-22n+16TakingLCMandtaking12common=nn+123n2+3n-4n-26=nn+1123n2-n-2
Now finding the value of S
S=nn+1243n2-3n+2n-2=nn+1243nn-1+2n-1=nn+1n-13n+224=nn2-13n+224
Therefore, the correct answer is option (A).
The sum of all the products of the first n natural numbers taken two at a time is