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Question Number 145516 by mathmax by abdo last updated on 05/Jul/21

$$\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{y}\right)+\mathrm{xy}\mathrm{for}\mathrm{all}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{fromR}$$$$\mathrm{and}\mathrm{f}\left(4\right)=10\mathrm{calculate}\mathrm{f}\left(1319\right)$$

Answered by Olaf_Thorendsen last updated on 05/Jul/21

$$f(x+y)=f\left(x\right)+f\left(y\right)+xy$$$$\mathrm{if}x=y=2$$$$f\left(4\right)=2f\left(2\right)+4\Rightarrow f\left(2\right)=\frac{10-4}{2}=3$$$$$$$$\mathrm{if}x=y=1$$$$f\left(2\right)=2f\left(1\right)+1\Rightarrow f\left(1\right)=\frac{3-1}{2}=1$$$$$$$$\mathrm{if}x=y=2^{n-1}$$$$f\left(2^n\right)=2f\left(2^{n-1}\right)+2^{2n-2}$$$$f\left(2^n\right)=2(2f\left(2^{n-2}\right)+2^{2n-4})+2^{2n-2}$$$$f\left(2^n\right)=2^{2}f\left(2^{n-2}\right)+2^{2n-2}+2^{2n-3}$$$$...$$$$f\left(2^n\right)=2^{n-2}f\left(2^2\right)+\underset{k=2}{\sum ^{n-1}}{2}^{2n-k}$$$$f\left(2^n\right)=10.2^{n-2}+2^{2n}\underset{k=2}{\sum ^{n-1}}{\left(\frac{1}{2}\right)}^k$$$$f\left(2^n\right)=10.2^{n-2}+2^{2n}\times \frac{1}{4}.\frac{1-{\left(\frac{1}{2}\right)}^{n-2}}{1-\frac{1}{2}}$$$$f\left(2^n\right)=10.2^{n-2}+2^{2n}\frac{1-{\left(\frac{1}{2}\right)}^{n-2}}{2}$$$$f\left(2^n\right)=10.2^{n-2}+2^{2n-1}-2^{n+1}$$$$$$$$1319=1024+256+32+4+2+1$$$$1319=2^{10}+2^8+2^5+4+2+1$$$$f\left(1024\right)=10.2^8+2^{19}-2^{11}=524800$$$$f\left(256\right)=10.2^6+2^{15}-2^9=32896$$$$f\left(32\right)=10.2^3+2^9-2^6=528$$$$$$$$f(1024+256)=f\left(1280\right)=f\left(1024\right)+f\left(256\right)+1024\times 256$$$$f\left(1280\right)=819840$$$$f(1280+32)=f\left(1312\right)=f\left(1280\right)+f\left(32\right)+1280\times 32$$$$f\left(1312\right)=861328$$$$f(1312+4)=f\left(1316\right)=f\left(1312\right)+f\left(4\right)+1312\times 4$$$$f\left(1316\right)=866586$$$$f(1316+2)=f\left(1318\right)=f\left(1316\right)+f\left(2\right)+1316\times 2$$$$f\left(1318\right)=869221$$$$f(1318+1)=f\left(1319\right)=f\left(1318\right)+f\left(1\right)+1318\times 1$$$$f\left(1319\right)=870540$$

Commented by mathmax by abdo last updated on 05/Jul/21

$$\mathrm{thanks}\mathrm{sir}\mathrm{olaf}$$

Answered by mathmax by abdo last updated on 05/Jul/21

$$\mathrm{we}\mathrm{have}\mathrm{f}(2+2)=\mathrm{f}\left(2\right)+\mathrm{f}\left(2\right)+4\Rightarrow \mathrm{f}\left(4\right)=2\mathrm{f}\left(2\right)+4\Rightarrow 2\mathrm{f}\left(2\right)=10-4=6$$$$\Rightarrow \mathrm{f}\left(2\right)=3$$$$\mathrm{f}(2+1)=\mathrm{f}\left(2\right)+\mathrm{f}\left(1\right)+2=5+\mathrm{f}\left(1\right)=\mathrm{f}\left(3\right)$$$$\mathrm{f}(1+1)=2\mathrm{f}\left(1\right)+1\Rightarrow 2\mathrm{f}\left(1\right)=\mathrm{f}\left(2\right)-\mathrm{f}\left(1\right)=2\Rightarrow \mathrm{f}\left(1\right)=1\Rightarrow \mathrm{f}\left(3\right)=6$$$$\mathrm{we}\mathrm{have}\mathrm{f}\left(1\right)=1,\mathrm{f}\left(2\right)=3=\frac{2(2+1)}{2},\mathrm{f}\left(3\right)=6=\frac{3(3+1)}{2}$$$$\mathrm{let}\mathrm{suppose}\mathrm{f}\left(\mathrm{n}\right)=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\Rightarrow \mathrm{f}(\mathrm{n}+1)=\mathrm{f}\left(\mathrm{n}\right)+\mathrm{f}\left(1\right)+\mathrm{n}$$$$=\frac{\mathrm{n}(\mathrm{n}+1)}{2}+1+\mathrm{n}=(\mathrm{n}+1)(\frac{\mathrm{n}}{2}+1)=\frac{(\mathrm{n}+1)(\mathrm{n}+2)}{2}\Rightarrow$$$$\mathrm{\forall }\mathrm{n}\in {\mathrm{N}}^{★}\mathrm{f}\left(\mathrm{n}\right)=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\Rightarrow$$$$\mathrm{f}\left(1319\right)=\frac{1}{2}\left(1319\right)\times \left(1320\right)=660\times 1319=66\times 13190=...$$

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