1 votes 1 votes If $a_n = 4^n + 6^n$ Find the value of $a_{40} \text { mod } 25$ Set Theory & Algebra binomial-distribution + – dd asked May 19, 2017 • retagged Jun 23, 2017 by Arjun dd 502 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 3 votes 3 votes correct me if i m wrong.... akash.dinkar12 answered May 19, 2017 • selected May 20, 2017 by srestha akash.dinkar12 comment Share Follow See all 3 Comments See all 3 3 Comments reply dd commented May 20, 2017 reply Follow Share HI, I have not checked every detail calculations of yours. But thanks for sharing it. Here is another way : $\begin{align*} &=\left [ 4^{40} + 6^{40} \right ] \mod 25 \\ &=\left [ \left ( 5-1 \right )^{40} + \left ( 5+1 \right )^{40} \right ] \mod 5^2 \\ &=\left [ \left \{ \binom{40}{0} \cdot 5^0 \cdot (-1)^{40} + \binom{40}{1} \cdot 5^1 \cdot (-1)^{39} \right \} + \left \{ \binom{40}{0} \cdot 5^0 \cdot (+1)^{40} + \binom{40}{1} \cdot 5^1 \cdot (+1)^{39} \right \} \right ] \mod 25 \\ &= \left [ 1 - 200 + 1 + 200 \right ] \mod 25 \\ &=2 \end{align*}$ 8 votes 8 votes Heisenberg commented May 30, 2017 reply Follow Share can u explain what principle did you use here? 0 votes 0 votes Shubhanshu commented Jul 18, 2017 reply Follow Share @Debashish Deka, here you have used binomial distribution, but why only for r = 0, 1? 0 votes 0 votes Please log in or register to add a comment.