## tính tổng s=1.3.4+2.5.7+…+n(2n+1)(3n+1)

Question

tính tổng s=1.3.4+2.5.7+…+n(2n+1)(3n+1)

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2 tháng 2021-07-24T19:45:32+00:00 1 Answers 1 views 0

$$\begin{array}{l} n\left( {2n + 1} \right).\left( {3n + 1} \right) = n.\left( {6{n^2} + 5n + 1} \right) = 6{n^3} + 5{n^2} + n\\ {1^3} + {2^3} + …. + {n^3} = {\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2}\\ {1^2} + {2^2} + {3^2} + ….. + {n^2} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\\ \Rightarrow S = 1.3.4 + 2.5.7 + …… + n.\left( {2n + 1} \right)\left( {3n + 1} \right)\\ = 6.\left( {{1^3} + {2^3} + …. + {n^3}} \right) + 5.\left( {{1^2} + {2^2} + ….. + {n^2}} \right) + \left( {1 + 2 + 3 + …. + n} \right)\\ = 6.{\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2} + 5.\frac{{n.\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{{n\left( {n + 1} \right)}}{2}\\ = \frac{{n\left( {n + 1} \right)}}{2}.\left[ {6.\frac{{n\left( {n + 1} \right)}}{2} + \frac{5}{3}.\left( {2n + 1} \right) + 1} \right]\\ = \frac{{n\left( {n + 1} \right)}}{2}.\left[ {3{n^2} + 3n + \frac{{10}}{3}n + \frac{8}{3}} \right]\\ = \frac{{n\left( {n + 1} \right)}}{2}\left[ {3{n^2} + \frac{{19}}{3}n + \frac{8}{3}} \right]\\ = \frac{{n\left( {n + 1} \right)\left( {9{n^2} + 19n + 8} \right)}}{6} \end{array}$$