Mathematics Tes...

For which positive integral values of n, is n! > 2ⁿ?

If P(n) = 2 + 4 + 6 + ......... + 2n, n – N then 6 P(m) = m(m + 1) + 2 
⇒ P(m + 1) = (m + 1) (m + 2) + 2 m – N, so we can conclude that P(n) = n(n+1) + 2 for

If α and β are the roots of the equation x² – x + 1 =0, then α²⁰⁰⁹ + β²⁰⁰⁹ =

Find all pairs of consecutive odd positive integer from the following options where both of which are smaller than 10 such that their sum is more than 11.