IPMAT Indore 2024Algebra > Hard(1−q)log(p)−(1−p)log(q)p−q\dfrac{(1-q)\log(p)-(1-p)\log(q)}{p-q}p−q(1−q)log(p)−(1−p)log(q)(1−q)log(q)−(1−p)log(p)p−q\dfrac{(1-q)\log(q)-(1-p)\log(p)}{p-q}p−q(1−q)log(q)−(1−p)log(p)(1−q)log(p)+(1−p)log(q)p−q\dfrac{(1-q)\log(p)+(1-p)\log(q)}{p-q}p−q(1−q)log(p)+(1−p)log(q)(1−q)log(q)+(1−p)log(p)p−q\dfrac{(1-q)\log(q)+(1-p)\log(p)}{p-q}p−q(1−q)log(q)+(1−p)log(p)✅ Correct Option: 2Related questions:If the sum of the first 212121 terms of the sequence: lnab,lnabb,lnab2,lnab2b,…\ln \frac{a}{b}, \ln \frac{a}{b \sqrt{b}}, \ln \frac{a}{b^{2}}, \ln \frac{a}{b^{2} \sqrt{b}}, \ldotslnba,lnbba,lnb2a,lnb2ba,… is lnambn\ln \frac{a^{m}}{b^{n}}lnbnam, then the value of m+nm+nm+n is \qquadGiven that 1+122+132+142+...=π261 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... = \frac{\pi^2}{6}1+221+321+421+...=6π2, the value of 1+132+152+172+...1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + ...1+321+521+721+... isThe sum up to 101010 terms of the series 1⋅3+5⋅7+9⋅11+...1 \cdot 3 + 5 \cdot 7 + 9 \cdot 11 + ...1⋅3+5⋅7+9⋅11+... is