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AP EAMCET · Maths · Sequences and Series

Let \(P(n): 1^2+2^2+3^2+\ldots+n^2\) \(=\frac{6(n-1)(n-2) \ldots(n-2020)+2 n^3+3 n^2+n}{6}\), for all \(n \in \mathbf{N}\). Then which of the following is correct?

  1. A \(P(n)\) is true for all \(n \in \mathrm{N}\)
  2. B \(P(n)\) is true for all \(h>2020\)
  3. C \(P(n)\) is true for all \(n \leq 2020\)
  4. D \(P(n)\) is not true for any \(n \in N\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(P(n)\) is true for all \(n \leq 2020\)

Step-by-step Solution

Detailed explanation

Given statement \(\begin{aligned} & P(n)=1^2+2^2+3^2+\ldots+n^2= \\ & \frac{6(n-1)(n-2) \ldots(n-2020)+2 n^3+3 n^2+n}{6} \end{aligned}\) \(\because\) We know that,…