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If `sum_(r=1)^(n)r(r+1)(2r+3)=an^(4)+bn^(3)+cn^(2)+dn+e`, then.A. `a+c=b+d`B. `e=0`C. `a,b-2//3,c-1` are in A.P.D. `c//a` is an integer |
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Answer» Correct Answer - A::B::C::D `sum_(r=1)^(n)r(r+1)(2r+3)=sum_(r=1)^(n)2r^(3)+5r^(2)+3r` `=2((n^(2)(n+1)^(2))/(4))+5((n(n+1)(2n+1))/(6))+3(n(n+1))/(2)` `=(3n^(2)(n+1)^(2)+5n(n+1)(2n+1)+9n(n+1))/(6)impliesa=(3)/(6)=(1)/(2)` `b=(16)/(6)=(8)/(3)` `c=(27)/(6)=(9)/(2)` `d=(14)/(6)=(7)/(3)` `e=0` |
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