For each n ∈ N, 10^{2n – 1} + 1 is divisible by _____________.

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Solution

For each n ∈ N,
Let P(n) : 10^2n–1 + 1
for n = 1
L.H.S = 10^2(1)–1 + 1
= 10¹ + 1
= 10 + 1
= 11
i.e P(1) = 11
Assume P(n) is true

for n = 2,
$P (2) : 10^{2 (2) - 1} + 1 = 10^{4 - 1} + 1 = 10^{3} + 1 = 1000 + 1 = 1001 P (2) = 11 (91)$
Similarly, assume that P(k) is divisible by 11.
$P (k + 1) : 10^{2 (k + 1) - 1} + 1 = 10^{2 k + 2 - 1} + 1 = 10^{2 k + 1} + 1 = 10^{(2 k - 1) + 2} + 1 = 10^{2 k - 1} \cdot 10^{2} + 1 [since 10^{2 k - 1} + 1 = 11 m i . e 10^{2 k - 1} = 11 m - 1] = (11 m - 1) 10^{2} + 1 = 11 m (10^{2}) - 10^{2} + 1 = 11 (100 m - 9) i . e P (n) is divisible by 11$

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