evaluate (summation of r=1 to r=n)sigma r = (2r-1) (2r+1) (2r+3)

Dear Student ,
Please find below the solution to the asked query :

r=1n2r-12r+12r+3=r=1n4r2-12r+3=r=1n8r3+12r2-2r-3=8nn+122+12nn+12n+16-2nn+12-3n=4n2n+12+2nn+12n+1-nn+1-3nFrom here, do calculation on your own for your practice to get the final answer
Hope this information will clear your doubts about the topic .
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