Prove by the method of mathematical induction that \[1\times3\times6+2\times4\times7+3\times5\times8+\cdots+n(n+2)(n+5)=\frac1{12}n(n+1)(3n^2+31n+74).\][6]
Show that \(\dfrac2{4r^2+8r+3}\) can be expressed as \(\dfrac A{2r+1}+\dfrac B{2r+3}\), where \(A\) and \(B\) are constants to be determined.[1]
The sum \(\displaystyle\sum_{r=1}^n\frac2{4r^2+8r+3}\) is denoted by \(S_n\). Find an expression for \(S_n\) in terms of \(n\).[4]
Find the smallest value of \(n\) for which \(S_n\) is within \(10^{-3}\) of the sum to infinity.[3]
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Answer:(b)(i) \(A=1,B=-1\) (ii) \(S_n=\frac13-\frac1{2n+3}\) (iii) \(499\).