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Question Number 202324 by hardmath last updated on 24/Dec/23

$$\mathrm{Find}:\frac{\frac{1}{2}+1+\frac{3}{2}+...+16}{\frac{1}{4}+\frac{2}{4}+\frac{3}{4}+...+8}$$

Answered by MATHEMATICSAM last updated on 24/Dec/23

$$\frac{\frac{1}{2}+1+\frac{3}{2}+....+16}{\frac{1}{4}+\frac{2}{4}+\frac{3}{4}+....+8}$$$$=\frac{\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{32}{2}}{\frac{1}{4}+\frac{2}{4}+\frac{3}{4}+\frac{4}{4}+....+\frac{32}{4}}$$$$=\frac{\frac{1}{2}[1+2+3+4+....+32]}{\frac{1}{4}[1+2+3+4+....+32]}$$$$=\frac{1}{2}\times 4$$$$=2$$

Commented by hardmath last updated on 24/Dec/23

$$\mathrm{thankyou}\mathrm{dear}\mathrm{professor}/$$

Answered by Rasheed.Sindhi last updated on 25/Dec/23

$$\frac{\frac{1}{2}+1+\frac{3}{2}+...+16}{\frac{1}{4}+\frac{2}{4}+\frac{3}{4}+...+8}$$$$Inordertokillthedenominators$$$$multiplynumerator\mathrm{\&}denominatorby4$$$$=\frac{4(\frac{1}{2}+1+\frac{3}{2}+...+16)}{4(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}+...+8)}$$$$=\frac{2\cancel{(1+2+3+...+32)}}{\cancel{(1+2+3+...+32)}}=2$$

Commented by MathematicalUser2357 last updated on 26/Dec/23

$$\mathrm{denominators}{\mathrm{isn}}^{\prime }\mathrm{t}\mathrm{humans}$$

Commented by Rasheed.Sindhi last updated on 26/Dec/23

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