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Question Number 129569 by zakirullah last updated on 16/Jan/21

Qi . using binomial theorem, prove that

(3^{2 n + 2} - 8 n - 9) is divisible by 64, where

n is positive integer .

Qii . By mathematical induction;

(\cos α + (n - 1) β) = \cos (α + \frac{(n - 1) β}{2}) \times \frac{\sin (n β / 2)}{\sin (β / 2)}

Answered by Dwaipayan Shikari last updated on 16/Jan/21

c o s α + c o s (α + β) + c o s (α + 2 β) + \dots + c o s (α + (n - 1) β)

= \frac{1}{2 s i n \frac{β}{2}} (s i n (α + \frac{β}{2}) - s i n (α - \frac{β}{2}) + s i n (α + \frac{3 β}{2}) - s i n (α + \frac{β}{2}) + \dots s i n (α + n β - \frac{β}{2}) - s i n (α + n β - \frac{3 β}{2}))

= \frac{1}{2 s i n \frac{β}{2}} (s i n (α + n β - \frac{β}{2}) - s i n (α - \frac{β}{2}))

= \frac{1}{s i n \frac{β}{2}} (c o s (α + \frac{(n - 1) β}{2}) s i n (\frac{n β}{2}))

Commented by zakirullah last updated on 16/Jan/21

thanks sir also find Qi ?

Answered by talminator2856791 last updated on 16/Jan/21

3^{2 n + 2} - 8 n - 9 = 9^{n + 1} - 8 n - 9

i . {(n + 1)}^{k} \equiv 1 (\mod n), k \in N

\Rightarrow 9^{n + 1} = {(8 + 1)}^{n + 1} \equiv 1 (\mod 8)

9^{n + 1} - 9 = 9 (9^{n} - 1) = 9 \cdot 8 \underset{k = 0}{\sum^{n - 1}} 9^{k}

from i ., deduce \underset{k = 0}{\sum^{n - 1}} 9^{k} \equiv n (\mod 8)

\Rightarrow 9 \cdot 8 \underset{k = 0}{\sum^{n - 1}} 9^{k} \equiv 9 \cdot 8 \cdot n (\mod 8)

9 \cdot 8 \underset{k = 0}{\sum^{n - 1}} 9^{k} \equiv 0 (\mod 8)

9 \cdot 8 \underset{k = 0}{\sum^{n - 1}} 9^{k} = 8 \cdot 8 \underset{k = 0}{\sum^{n - 1}} 9^{k} + 8 \cdot \underset{k = 0}{\sum^{n - 1}} 9^{k}

8 \underset{k = 0}{\sum^{n - 1}} 9^{k} \equiv 0 (\mod 8)

\Rightarrow 8 \underset{k = 0}{\sum^{n - 1}} 9^{k} \equiv 8 n (\mod 64)

\Rightarrow 3^{2 n + 2} - 9 \equiv 8 n (\mod 64)

3^{2 n + 2} - 8 n - 9 \equiv 0 (\mod 64)

Commented by zakirullah last updated on 16/Jan/21

you are the great man .