If 1^{3}+2^{3}+cdots+9^{3}=2025, then the val | vedclass

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(C) Given that $1^{3}+2^{3}+\cdots+9^{3}=2025$. We need to find the value of $(0.11)^{3}+(0.22)^{3}+\cdots+(0.99)^{3}$. This can be written as $(0.11

Vedclass · Accepted Answer

$2.695$

Vedclass · Answer

(C) Given that $1^{3}+2^{3}+\cdots+9^{3}=2025$. We need to find the value of $(0.11)^{3}+(0.22)^{3}+\cdots+(0.99)^{3}$. This can be written as $(0.11 	imes 1)^{3} + (0.11 	imes 2)^{3} + \cdots + (0.11 	imes 9)^{3}$. Factoring out $(0.11)^{3}$,we get $(0.11)^{3} 	imes (1^{3} + 2^{3} + \cdots + 9^{3})$. Substituting the given sum,we have $(0.11)^{3} 	imes 2025$. Since $(0.11)^{3} = 0.001331$,the expression becomes $0.001331 	imes 2025$. Calculating the product: $0.001331 	imes 2025 = 2.695$.

If $1^{3}+2^{3}+\cdots+9^{3}=2025,$ then the value of $(0.11)^{3}+(0.22)^{3}+\cdots+(0.99)^{3}$ is close to

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