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Question Number 40254 by scientist last updated on 17/Jul/18

sum of this series: 1.3.5+3.5.7+5.7.9+...up to n terms

$${sum}\:{of}\:{this}\:{series}: \\ $$$$\mathrm{1}.\mathrm{3}.\mathrm{5}+\mathrm{3}.\mathrm{5}.\mathrm{7}+\mathrm{5}.\mathrm{7}.\mathrm{9}+...{up}\:{to}\:{n}\:{terms} \\ $$

Commented by maxmathsup by imad last updated on 18/Jul/18

Σ_(k=0) ^n (2k+1)(2k+3)(2k+5)=1.3.5 +3.5.7 +5.7.9+....+(2n+1)(2n+3)(2n+5) =S_n S_n = Σ_(k=0) ^n (4k^2 +6k +2k+3)(2k+5) =Σ_(k=0) ^n (4k^2 +8k+3)(2k+5) =Σ_(k=0) ^n (8k^3 +20k^2 +16k^2 +40k +6k +15) =Σ_(k=0) ^n (8k^3 +32k^2 +46k +15) =8Σ_(k=0) ^n k^3 +32 Σ_(k=0) ^n k^2 +46 Σ_(k=0) ^n k +15 Σ_(k=0) ^n (1) =8 ((n^2 (n+1)^2 )/4) +32 ((n(n+1)(2n+1))/6) +46 ((n(n+1))/2) +15n =2n^2 (n+1)^2 + ((16)/3) n(n+1)(2n+1) +23n(n+1) +15n .

$$\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{3}\right)\left(\mathrm{2}{k}+\mathrm{5}\right)=\mathrm{1}.\mathrm{3}.\mathrm{5}\:+\mathrm{3}.\mathrm{5}.\mathrm{7}\:+\mathrm{5}.\mathrm{7}.\mathrm{9}+....+\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\mathrm{2}{n}+\mathrm{5}\right) \\ $$$$={S}_{{n}} \\ $$$${S}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}} \left(\mathrm{4}{k}^{\mathrm{2}} \:+\mathrm{6}{k}\:+\mathrm{2}{k}+\mathrm{3}\right)\left(\mathrm{2}{k}+\mathrm{5}\right) \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \left(\mathrm{4}{k}^{\mathrm{2}} \:+\mathrm{8}{k}+\mathrm{3}\right)\left(\mathrm{2}{k}+\mathrm{5}\right) \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(\mathrm{8}{k}^{\mathrm{3}} \:+\mathrm{20}{k}^{\mathrm{2}} \:+\mathrm{16}{k}^{\mathrm{2}} \:+\mathrm{40}{k}\:+\mathrm{6}{k}\:+\mathrm{15}\right) \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \left(\mathrm{8}{k}^{\mathrm{3}} \:+\mathrm{32}{k}^{\mathrm{2}} \:+\mathrm{46}{k}\:+\mathrm{15}\right) \\ $$$$=\mathrm{8}\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}^{\mathrm{3}} \:\:+\mathrm{32}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}^{\mathrm{2}} \:+\mathrm{46}\:\sum_{{k}=\mathrm{0}} ^{{n}} {k}\:+\mathrm{15}\:\sum_{{k}=\mathrm{0}} ^{{n}} \left(\mathrm{1}\right) \\ $$$$=\mathrm{8}\:\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\:\:+\mathrm{32}\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\:+\mathrm{46}\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:\:+\mathrm{15}{n} \\ $$$$=\mathrm{2}{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} \:\:+\:\frac{\mathrm{16}}{\mathrm{3}}\:{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\:+\mathrm{23}{n}\left({n}+\mathrm{1}\right)\:\:+\mathrm{15}{n}\:. \\ $$

Commented by scientist last updated on 18/Jul/18

but can′t we use Tn=n(n+2)(n+4)

$${but}\:{can}'{t}\:{we}\:{use}\:\:{Tn}={n}\left({n}+\mathrm{2}\right)\left({n}+\mathrm{4}\right) \\ $$

Commented by maxmathsup by imad last updated on 18/Jul/18

no look sir Σ_(k=1) ^n T_k =Σ_(k=1) ^n k(k+2)(k+4)= 1.3.4 +2.4.6 +... and this sum is totally differnt from the sum given in the question...

$${no}\:\:{look}\:{sir}\:\:\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{T}_{{k}} =\sum_{{k}=\mathrm{1}} ^{{n}} {k}\left({k}+\mathrm{2}\right)\left({k}+\mathrm{4}\right)=\:\mathrm{1}.\mathrm{3}.\mathrm{4}\:\:+\mathrm{2}.\mathrm{4}.\mathrm{6}\:+... \\ $$$${and}\:{this}\:{sum}\:{is}\:{totally}\:{differnt}\:{from}\:{the}\:{sum}\:{given}\:{in}\:{the}\:{question}... \\ $$

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