Question

a. 1 x 2 + 2 × 3 + 3 × 4 + ... 9 × 10 =
b. 111 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 118 + 117 + ... + 112 + 111 =
va rog urgent dacă nu puteți sa le faceți pe amândouă măcar una va rog

Answer 1

[tex]a)1\cdot2+2\cdot3+3\cdot4+...+9\cdot10=\\ 1(1+1)+2(2+1)+3(3+1)+...+9(9+1)=\\ 1+1+2^2+2+3^2+3+..+9^2+9=\\ (1+2+3+...+9)+(1+2^2+3^2+...+9^2)=\\ Pentru\ prima\ paranteza\ folosim\ formula\ lui\ Gauss.\\ Pentru\ a\ doua\ paranteza\ folosim\ formula:\\ \boxed{1+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}}\\ Revenind: \frac{9\cdot10}{2}+\frac{9\cdot10\cdot19}{6}=\\ 45+285=\boxed{330}\\ [/tex]
[tex]b. 111+112+113+..+118+118+117+...+111=\\ 2\cdot111+2\cdot112+2\cdot113+...+2\cdot118=\\ 2(111+112+113+....+118)=\\ 2(1+2+3+...+118-1-2-...-110)=\\ 2(\frac{118\cdot119}{2}-\frac{110\cdot111}{2})=\\ 118\cdot119-110\cdot111=\\ 14042-12210=\boxed{1832}[/tex]

Answer 2

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