InterviewSolution
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InterviewSolution
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Find the nth term and sum to n terms of the series `12+40+90+168+280+432+"...."`. |
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Answer» Let `S_(n)=12+40+90+168+280+432+"....",` then 1st order differences are `28,50,78,112,152,"...."(i.e.Deltat_(1),Deltat_(2),Deltat_(3),".....")` and 2nd order differences are `22,28,34,40,"...."(i.e.Deltat_(1)^(2),Deltat_(2)^(2),Deltat_(3)^(2),".....")` and 3rd order dofferences are `6,6,6,6,"...."(i.e.Deltat_(1)^(3),Deltat_(2)^(3),Deltat_(3)^(3),".....")` and 4th order differences are `0,0,0,0,"...."(i.e.Deltat_(1)^(4),Deltat_(2)^(4),Deltat_(3)^(4),".....")` `therefore t_(n)=12*^(n-1)C_(0)+28*^(n-1)C_(1)+22*^(n-1)C_(2)+6*^(n-1)C_(3)` `=12+28(n-1)+(22(n-1)(n-2))/(2)+(6(n-1)(n-2)(n-3))/(1*2*3)` `=n^(3)+5n^(2)+6n` and `S_(n)=12*^(n)C_(1)+28*^(n)C_(2)+22*^(n)C_(3)+6*^(n)C_(4)` `=12n+(28n(n-1))/(2)+(22n(n-1)(n-2))/(1*2*3)+(6*n(n-1)(n-2)(n-3))/(1*2*3*4)` `=(n)/(12)(n+1)(3n^(2)+23n+46)`. |
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