Given:
\(x^2 - x - 1 = 0 \implies \alpha, \beta \text{ are roots.}\)
The relation between \(\alpha\) and \(\beta\) is:
\(\alpha^2 = \alpha + 1, \quad \beta^2 = \beta + 1.\)
The sequence \(S_n\) is defined as:
\(S_n = 2023\alpha^n + 2024\beta^n.\)
Using the recurrence relation for the roots:
\(S_{n+2} = S_{n+1} + S_n.\)
Applying this for \(n = 10\):
\(S_{12} = S_{11} + S_{10}.\)
The Correct answer is: \( 2S_{11} = S_{12} + S_{10} \)
If \(\sum\)\(_{r=1}^n T_r\) = \(\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}\) , then \( \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r} \) is equal to :
Let $ f(x) = \begin{cases} (1+ax)^{1/x} & , x<0 \\1+b & , x = 0 \\\frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} & , x>0 \end{cases} $ be continuous at x = 0. Then $ e^a bc $ is equal to
Total number of nucleophiles from the following is: \(\text{NH}_3, PhSH, (H_3C_2S)_2, H_2C = CH_2, OH−, H_3O+, (CH_3)_2CO, NCH_3\)