Q79 / 160 · AP EAMCET 2020 (21 Sep Shift 1) · Mathematics

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?

  • A
    \(P(n)\) is true for all \(n \in \mathrm{N}\)
  • B
    \(P(n)\) is true for all \(h>2020\)
  • Correct answer
    C
    \(P(n)\) is true for all \(n \leq 2020\)
  • D
    \(P(n)\) is not true for any \(n \in N\)

Solution

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, \(\begin{aligned} & 1^2+2^2+3^2+\ldots+n^2=\frac{n(n+1)(2 n+1)}{6} \\ & \therefore \quad \frac{6(n-1)(n-2) \ldots(n-2020)+2 n^3+3 n^2+n}{6} \\ & \quad=(n-1)(n-2) \ldots(n-2020)+\frac{n(n+1)(2 n+1)}{6} \end{aligned}\) will be \(\frac{n(n+1)(2 n+1)}{6}\) if \(n=1,2,3, \ldots ., 2020\) only. Therefore, \(P(n)\) is true for all \(n \leq 2020\). Hence, option (c) is correct.

Hence, the correct answer is option: 'C'.

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