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 Solution Prove that 2 + 5 + 10 + 17 + ... n terms = n/6 (2n^2 + 3n + 7)Solution:Your problem -> Prove that 2 + 5 + 10 + 17 + ... n terms = n/6 (2n^2 + 3n + 7)L.H.S. = 2 + 5 + 10 + 17 + ... n terms= sum [ 1 + n^2 ]= sum 1 + sum n^2= n + (n (n + 1) (2n + 1))/6= n/6 [ 6 + (2n^2 + 3n + 1) ]= n/6 (2n^2 + 3n + 7)= R.H.S. (Proved) Solution provided by AtoZmath.com Any wrong solution, solution improvement, feedback then Submit Here Want to know about AtoZmath.com and me