Let r_k=\frac\int_0^1(1-x^7)^k d x\int_0^1(1-x^7)^k+1 d x, k \in \mathbbN . Then the value of \sum_k=1^10 \frac17(r_k-1) is equal to \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2024 · 6 Apr · Shift 1 · Q58 with Solution

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Correct answer: 65

Let $I_k=\int_0^1 (1-x^7)^k\,dx.$ Then $r_k=\frac{I_k}{I_{k+1}}.$ We need to find $\sum_{k=1}^{10}\frac{1}{7(r_k-1)}.$
First, derive a relation between $I_k$ and $I_{k+1}$ .

Use the substitution $t=x^7 \implies x=t^{1/7}, \quad dx=\frac{1}{7}t^{-6/7}dt.$ So,

=\frac{1}{7}B\left(\frac17,k+1\right).$$ Using the Beta-function identity $$B(a,b+1)=\frac{b}{a+b}B(a,b),$$ with $a=\frac17$ and $b=k+1$, we get $$B\left(\frac17,k+2\right)=\frac{k+1}{k+1+\frac17}B\left(\frac17,k+1\right).$$ Hence $$I_{k+1}=\frac{k+1}{k+\frac87}I_k.$$ Therefore, $$r_k=\frac{I_k}{I_{k+1}}=\frac{k+\frac87}{k+1}= rac{7k+8}{7k+7}.$$ 3. Now compute $r_k-1$: $$r_k-1=\frac{7k+8}{7k+7}-1=\frac{1}{7k+7}=\frac{1}{7(k+1)}.$$ Thus, $$7(r_k-1)=\frac{1}{k+1}.$$ So, $$\frac{1}{7(r_k-1)}=k+1.$$ 4. Therefore, $$\sum_{k=1}^{10}\frac{1}{7(r_k-1)}=\sum_{k=1}^{10}(k+1)=2+3+\cdots+11.$$ This is an arithmetic series with 10 terms: $$\sum_{k=1}^{10}(k+1)=\sum_{k=1}^{10}k+\sum_{k=1}^{10}1=\frac{10\cdot 11}{2}+10=55+10=65.$$ 5. Final answer: $$\boxed{65}$$ The derived answer matches the stored correct answer.

View written solutionFree

Correct answer: 65

Let $I_k=\int_0^1 (1-x^7)^k\,dx.$ Then $r_k=\frac{I_k}{I_{k+1}}.$ We need to find $\sum_{k=1}^{10}\frac{1}{7(r_k-1)}.$
First, derive a relation between $I_k$ and $I_{k+1}$ .

Use the substitution $t=x^7 \implies x=t^{1/7}, \quad dx=\frac{1}{7}t^{-6/7}dt.$ So,

=\frac{1}{7}B\left(\frac17,k+1\right).$$ Using the Beta-function identity $$B(a,b+1)=\frac{b}{a+b}B(a,b),$$ with $a=\frac17$ and $b=k+1$, we get $$B\left(\frac17,k+2\right)=\frac{k+1}{k+1+\frac17}B\left(\frac17,k+1\right).$$ Hence $$I_{k+1}=\frac{k+1}{k+\frac87}I_k.$$ Therefore, $$r_k=\frac{I_k}{I_{k+1}}=\frac{k+\frac87}{k+1}= rac{7k+8}{7k+7}.$$ 3. Now compute $r_k-1$: $$r_k-1=\frac{7k+8}{7k+7}-1=\frac{1}{7k+7}=\frac{1}{7(k+1)}.$$ Thus, $$7(r_k-1)=\frac{1}{k+1}.$$ So, $$\frac{1}{7(r_k-1)}=k+1.$$ 4. Therefore, $$\sum_{k=1}^{10}\frac{1}{7(r_k-1)}=\sum_{k=1}^{10}(k+1)=2+3+\cdots+11.$$ This is an arithmetic series with 10 terms: $$\sum_{k=1}^{10}(k+1)=\sum_{k=1}^{10}k+\sum_{k=1}^{10}1=\frac{10\cdot 11}{2}+10=55+10=65.$$ 5. Final answer: $$\boxed{65}$$ The derived answer matches the stored correct answer.