1^(3)+2^(3)+3^(3)+4^(3)+5^(3)+.....+49^(3)=.... Question from @fidyachykita11

October 2018 2 20 Report

1^(3)+2^(3)+3^(3)+4^(3)+5^(3)+.....+49^(3)=

ImmanuelSinaga $\bold{berlaku~:} \\ \\ \boxed{\bold{(a)^3+(a+1)^3+(a+2)^3+.......~=~(a+(a+1)+(a+2)+......)^2}} \\ \\ \bold{Maka~:} \\ 1^3+2^3+4^3+5^3+.......+49^3~~~=~~~~(1+2+3+4+5.......+49)^2 \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=~~~~(\bold{S_{49}})^2 \\ \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=~~~~(\bold{\displaystyle\frac{49}{2}~~\times~~(1+49)})^2 \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=~~~~\bold{(1225)^2~~~=~~~1.500.625}$

untuk mencari 1 + 2+ 3 + ........ + 49 , menggunakan deret aritmatika , dengan :
beda = 1
suku awal = 1

1 votes Thanks 1

Syubbana 1^3 + 2^3 +3^3 + ... + 49^3 =

(1+2+3+...+49)^2 =

1225^2 =

1.500.625

rumusnya a^3 + b^3 +c^3 + .... + z^3 = (a+b+c+ ... +z)^2

1 votes Thanks 0

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