AP EAMCET · Maths · Sequences and Series
For all \(\mathrm{n} \in \mathrm{N}\), if \(1^3+2^3+3^3+\ldots. .+\mathrm{n}^3>\mathrm{x}\), then a value of x among the following is
- A \(\frac{n^2}{4}\)
- B \(\mathrm{n}^2\)
- C \(\mathrm{n}^4\)
- D \(\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{n^2}{4}\)
Step-by-step Solution
Detailed explanation
\(1^3+2^3+3^3+\ldots.+\mathrm{n}^3 = \frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}\) Check \(\mathrm{x}=\frac{\mathrm{n}^2}{4}\): \(\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4} > \frac{\mathrm{n}^2}{4}\) \((\mathrm{n}+1)^2 > 1\) \(\mathrm{n}+1 > 1 \implies \mathrm{n} > 0\) This is true for…
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