1 + 2 + 22 + … 2n = 2n+1 – 1 for all natural numbers n.
According to the question,
P(n) is 1 + 2 + 22 + … 2n = 2n+1 – 1.
So, substituting different values for n, we get,
P(0) = 1 = 20+1 − 1 Which is true.
P(1) = 1 + 2 = 3 = 21+1 − 1 Which is true.
P(2) = 1 + 2 + 22 = 7 = 22+1 − 1 Which is true.
P(3) = 1 + 2 + 22 + 23 = 15 = 23+1 − 1 Which is true.
Let P(k) = 1 + 2 + 22 + … 2k = 2k+1 – 1 be true;
So, we get,
⇒ P(k+1) is 1 + 2 + 22 + … 2k + 2k+1 = 2k+1 – 1 + 2k+1
= 2×2k+1 – 1
= 2(k+1)+1 – 1
⇒ P(k+1) is true when P(k) is true.
Therefore, by Mathematical Induction,
1 + 2 + 22 + … 2n = 2n+1 – 1 is true for all natural numbers n.
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