SHOW THAT : 2^(2N) – 3N -1 is divisible by 9 FOR ALL N>=1
This question delves into deep philosophical and existential speculation. The last thought of the last conscious being in the universe before existence ends could encompass various profound themes: Reflection on Existence: A contemplation on the nature of existence itself—what it meant to be, to livRead more
This question delves into deep philosophical and existential speculation. The last thought of the last conscious being in the universe before existence ends could encompass various profound themes:
- Reflection on Existence: A contemplation on the nature of existence itself—what it meant to be, to live, and to perceive.
- Loneliness or Unity: A feeling of ultimate solitude or perhaps a sense of unity with everything that has ever existed.
- Gratitude or Regret: A final expression of gratitude for having experienced life or regret for unfinished endeavors or lost opportunities.
- Acceptance or Resistance: A thought of acceptance, embracing the end as a natural part of existence, or resistance, a desire for life and consciousness to continue.
- Memory or Forgetting: A recollection of memories, the summation of life’s experiences, or the fading away into oblivion, as if even memory itself ceases to hold meaning.
- Wonder or Understanding: A profound wonder about the mysteries of the universe or a serene understanding, a moment of ultimate clarity.
Proof by Mathematical Induction: Base Case: Let \( n = 1 \) \[ 2^{2 \cdot 1} - 3 \cdot 1 - 1 = 2^2 - 3 - 1 = 4 - 3 - 1 = 0 \] \[ \Rightarrow 9 \mid 0 \quad \text{(True)} \] Inductive Hypothesis: Assume that for some \( n = k \), the expression is divisible by 9: \[ 2^{2k} - 3k - 1 \equiv 0 \pmod{9}Read more
Proof by Mathematical Induction:
Base Case: Let \( n = 1 \)
\[
2^{2 \cdot 1} – 3 \cdot 1 – 1 = 2^2 – 3 – 1 = 4 – 3 – 1 = 0
\]
\[
\Rightarrow 9 \mid 0 \quad \text{(True)}
\]
Inductive Hypothesis: Assume that for some \( n = k \), the expression is divisible by 9:
\[
2^{2k} – 3k – 1 \equiv 0 \pmod{9}
\]
That is,
\[
2^{2k} – 3k – 1 = 9m \quad \text{for some integer } m
\]
To Prove: \( 2^{2(k+1)} – 3(k+1) – 1 \) is also divisible by 9.
LHS:
\[
2^{2(k+1)} – 3(k+1) – 1 = 2^{2k + 2} – 3k – 3 – 1
\]
\[
= 4 \cdot 2^{2k} – 3k – 4
\]
Now subtract the inductive hypothesis:
\[
[4 \cdot 2^{2k} – 3k – 4] – [2^{2k} – 3k – 1] = 3 \cdot 2^{2k} – 3
\]
\[
= 3(2^{2k} – 1)
\]
Now, observe that \( 2^{2k} \equiv 1 \pmod{3} \), so \( 2^{2k} – 1 \equiv 0 \pmod{3} \)
Hence, \( 3(2^{2k} – 1) \equiv 0 \pmod{9} \)
Therefore, the expression for \( n = k+1 \) is also divisible by 9.
Conclusion:
\[
See less\boxed{2^{2n} – 3n – 1 \text{ is divisible by } 9 \text{ for all } n \geq 1}
\]