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Question Number 179773 by Acem last updated on 02/Nov/22

What′s the sum of the odd numbers between 2313 and 4718

$${What}'{s}\:{the}\:{sum}\:{of}\:{the}\:{odd}\:{numbers} \\ $$$$\:{between}\:\mathrm{2313}\:{and}\:\mathrm{4718} \\ $$

Commented by CElcedricjunior last updated on 05/Nov/22

calculons la somme des nombres comprises entre 2313 et 4718 ⇔2314+2315+......+4717 =(2313+1)+(2313+2)+......(2313+2404) =2313×2404+((2404×2405)/2) =2313×2404+1202×2405 =8.451.262 d′ou^� Σ_(k=2314) ^(4717) k=8.451.262

$$\boldsymbol{{calculons}}\:\boldsymbol{{la}}\:\boldsymbol{{somme}}\:\boldsymbol{{des}}\:\boldsymbol{{nombres}} \\ $$$$\boldsymbol{{comprises}}\:\boldsymbol{{entre}}\:\mathrm{2313}\:\boldsymbol{{et}}\:\mathrm{4718} \\ $$$$\Leftrightarrow\mathrm{2314}+\mathrm{2315}+......+\mathrm{4717} \\ $$$$=\left(\mathrm{2313}+\mathrm{1}\right)+\left(\mathrm{2313}+\mathrm{2}\right)+......\left(\mathrm{2313}+\mathrm{2404}\right) \\ $$$$=\mathrm{2313}×\mathrm{2404}+\frac{\mathrm{2404}×\mathrm{2405}}{\mathrm{2}} \\ $$$$=\mathrm{2313}×\mathrm{2404}+\mathrm{1202}×\mathrm{2405} \\ $$$$=\mathrm{8}.\mathrm{451}.\mathrm{262} \\ $$$${d}'\boldsymbol{{o}}\grave {\boldsymbol{{u}}}\:\underset{\boldsymbol{{k}}=\mathrm{2314}} {\overset{\mathrm{4717}} {\sum}}\boldsymbol{{k}}=\mathrm{8}.\mathrm{451}.\mathrm{262} \\ $$

Answered by Rasheed.Sindhi last updated on 02/Nov/22

Sum of odd numbers between 2313 and 4718( exclusively) 2315+2317+...+4717 a=2315,l=4717, d=2, l=a_n =a+(n−1)d=2315+2(n−1)=4717 n=(4717−2315+2)/2=1202 determinant (((S_n =(n/2)(a+l)))) S_(1202) =((1202)/2)(2315+4717)=4226232

$${Sum}\:{of}\:{odd}\:{numbers} \\ $$$$\:{between}\:\mathrm{2313}\:{and}\:\mathrm{4718}\left(\:{exclusively}\right) \\ $$$$\mathrm{2315}+\mathrm{2317}+...+\mathrm{4717} \\ $$$${a}=\mathrm{2315},{l}=\mathrm{4717},\:{d}=\mathrm{2}, \\ $$$$\:{l}={a}_{{n}} ={a}+\left({n}−\mathrm{1}\right){d}=\mathrm{2315}+\mathrm{2}\left({n}−\mathrm{1}\right)=\mathrm{4717} \\ $$$${n}=\left(\mathrm{4717}−\mathrm{2315}+\mathrm{2}\right)/\mathrm{2}=\mathrm{1202} \\ $$$$\begin{array}{|c|}{{S}_{{n}} =\frac{{n}}{\mathrm{2}}\left({a}+{l}\right)}\\\hline\end{array} \\ $$$${S}_{\mathrm{1202}} =\frac{\mathrm{1202}}{\mathrm{2}}\left(\mathrm{2315}+\mathrm{4717}\right)=\mathrm{4226232} \\ $$

Commented by Acem last updated on 02/Nov/22

Thanks for your efforts and writing details For me i use the sum till 1 and subtract Sum= ((n_1 /2))^2 − ((n_2 /2))^2 ; n_i even = (((4718)/2))^2 − (((2314)/2))^2 = 4 226 232

$${Thanks}\:{for}\:{your}\:{efforts}\:{and}\:{writing}\:{details} \\ $$$${For}\:{me}\:{i}\:{use}\:{the}\:{sum}\:{till}\:\mathrm{1}\:{and}\:{subtract} \\ $$$$\:{Sum}=\:\left(\frac{{n}_{\mathrm{1}} }{\mathrm{2}}\right)^{\mathrm{2}} −\:\left(\frac{{n}_{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} \:;\:{n}_{{i}} \:{even} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\left(\frac{\mathrm{4718}}{\mathrm{2}}\right)^{\mathrm{2}} \:−\:\left(\frac{\mathrm{2314}}{\mathrm{2}}\right)^{\mathrm{2}} \:=\:\mathrm{4}\:\mathrm{226}\:\mathrm{232} \\ $$$$ \\ $$

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