Let r_k= _0^1 0.05em(1-x^7)^k dx _0^1 0.05em(1-x^7)^k+1 dx ,k ∈N. Then the value of _k=1^10 0.05em 17(r_k-1) is equal to________

Question

Accepted Answer

Correct answer: 65. (65) I_K= 1 (1-x^7)^K dxI_K= ((1-x^7)^Kx )_0^1+7K _0^1 0.05em(1-x^7)^K-1x^6 xdxI_K=-7K _0^1 0.05em(1-x^7)^K-1((1-x^7)-1) dxI_K=-7KI_K+7KI_K-1 ⇒ I_KI_K+1= 7K+87K+7r_K= 7K+87 K+7r_K-1= 17 ( K+1 ) ⇒7(r_K-1)= 1 K+1 _K=1^10 ( K+1 )=11 (6 )-1=65

$Let r_{k} = \frac{\int _{0}^{1} ( 1 - x ^{7} ) ^{k} dx}{\int _{0}^{1} ( 1 - x ^{7} ) ^{k + 1} dx}, k \in N . Then the value of \sum_{k = 1}^{10} \frac{1}{7 ( r _{k} - 1 )} is equal to________$

Solution

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