JEE Main 2024 — 6 April, Shift 1 — Question 23

$\quad I_{K}=\int 1 .\left(1-x^{7}\right)^{\mathrm{K}} d x$

$$I_{K}=\left.\left(1-x^{7}\right)^{K} x\right|_{0} ^{1}+7 K \int_{0}^{1}\left(1-x^{7}\right)^{K-1} x^{6} \cdot x d x$$

$I_{K}=-7 K \int_{0}^{1}\left(1-x^{7}\right)^{K-1}\left(\left(1-x^{7}\right)-1\right) d x$

$\mathrm{I}_{\mathrm{K}}=-7 \mathrm{~K} \mathrm{I}_{\mathrm{K}}+7 \mathrm{KI}_{\mathrm{K}-1}$

$\Rightarrow \frac{\mathrm{I}_{\mathrm{K}}}{\mathrm{I}_{\mathrm{K}+1}}=\frac{7 \mathrm{~K}+8}{7 \mathrm{~K}+7}$

$\mathrm{r}_{\mathrm{K}}=\frac{7 \mathrm{~K}+8}{7 \mathrm{~K}+7}$

$\mathrm{r}_{\mathrm{K}}-1=\frac{1}{7(\mathrm{~K}+1)}$

$\Rightarrow 7\left(\mathrm{r}_{\mathrm{K}}-1\right)=\frac{1}{\mathrm{~K}+1}$

$\sum_{K=1}^{10}(K+1)=11(6)-1=65$