For each n ∈ N,

Question:

For each n ∈ N, 102n – 1 + 1 is divisible by _____________.

Solution:

For each n ∈ N,

Let P(n) : 102n–1 + 1

for n = 1

L.H.S = 102(1)–1 + 1

= 101 + 1

= 10 + 1

= 11

i.e P(1) = 11

Assume P(n) is true

for = 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.

$P(k+1): 10^{2(k+1)-1}+1$

$=10^{2 k+2-1}+1$

$=10^{2 k+1}+1$

$=10^{(2 k-1)+2}+1$

$=10^{2 k-1} \cdot 10^{2}+1 \quad\left[\right.$ since $10^{2 k-1}+1=11 \mathrm{~m}$ i. e $\left.10^{2 k-1}=11 \mathrm{~m}-1\right]$

$=(11 m-1) 10^{2}+1$

$=11 m\left(10^{2}\right)-10^{2}+1$

$=11(100 m-9)$ i. e $P(n)$ is divisible by 11