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

Find: (((1/2) + 1 + (3/2) + ... + 16)/((1/4) + (2/4) + (3/4) + ... + 8))

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

Answered by MATHEMATICSAM last updated on 24/Dec/23

(((1/2) + 1 + (3/2) + .... + 16)/((1/4) + (2/4) + (3/4) + .... + 8)) = (((1/2) + (2/2) + (3/2) + (4/2) + .... + ((32)/2))/((1/4) + (2/4) + (3/4) + (4/4) + .... + ((32)/4))) = (((1/2)[1 + 2 + 3 + 4 + .... + 32])/((1/4)[1 + 2 + 3 + 4 + .... + 32])) = (1/2) × 4 = 2

$$\frac{\frac{\mathrm{1}}{\mathrm{2}}\:+\:\mathrm{1}\:+\:\frac{\mathrm{3}}{\mathrm{2}}\:+\:....\:+\:\mathrm{16}}{\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{2}}{\mathrm{4}}\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:+\:....\:+\:\mathrm{8}} \\ $$$$=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{2}}{\mathrm{2}}\:+\:\frac{\mathrm{3}}{\mathrm{2}}\:+\:\frac{\mathrm{4}}{\mathrm{2}}\:+\:....\:+\:\frac{\mathrm{32}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{2}}{\mathrm{4}}\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:+\:\frac{\mathrm{4}}{\mathrm{4}}\:+\:....\:+\:\frac{\mathrm{32}}{\mathrm{4}}} \\ $$$$=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:\mathrm{4}\:+\:....\:+\:\mathrm{32}\right]}{\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:\mathrm{4}\:+\:....\:+\:\mathrm{32}\right]} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:×\:\mathrm{4} \\ $$$$=\:\mathrm{2}\: \\ $$

Commented by hardmath last updated on 24/Dec/23

thankyou dear professor/

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

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

(((1/2) + 1 + (3/2) + ... + 16)/((1/4) + (2/4) + (3/4) + ... + 8)) In order to kill the denominators multiply numerator & denominator by 4 =((4((1/2) + 1 + (3/2) + ... + 16))/(4((1/4) + (2/4) + (3/4) + ... + 8))) =((2(1+2+3+...+32))/((1+2+3+...+32)))=2

$$\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\:+\:\mathrm{1}\:+\:\frac{\mathrm{3}}{\mathrm{2}}\:+\:...\:+\:\mathrm{16}}{\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{2}}{\mathrm{4}}\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:+\:...\:+\:\mathrm{8}} \\ $$$${In}\:{order}\:{to}\:{kill}\:{the}\:{denominators} \\ $$$${multiply}\:{numerator}\:\&\:{denominator}\:{by}\:\mathrm{4} \\ $$$$\:=\frac{\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{2}}\:+\:\mathrm{1}\:+\:\frac{\mathrm{3}}{\mathrm{2}}\:+\:...\:+\:\mathrm{16}\right)}{\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{2}}{\mathrm{4}}\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:+\:...\:+\:\mathrm{8}\right)} \\ $$$$=\frac{\mathrm{2}\cancel{\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+...+\mathrm{32}\right)}}{\cancel{\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+...+\mathrm{32}\right)}}=\mathrm{2} \\ $$

Commented by MathematicalUser2357 last updated on 26/Dec/23

denominators isn′t humans

$$\mathrm{denominators}\:\mathrm{isn}'\mathrm{t}\:\mathrm{humans} \\ $$

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

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