Question

please please please answer those two questions....

..if you can..don't spam or I will immediately report you...guys it's urgent need...so please be helpful..don't..attempt only to get points...please I need correct✅ answers it's urgent​

Answer 1

Answer:

Let n = 1

$3 = \frac{1}{27} ( {10}^{1 + 1} - 9 - 10) = 3$

So it is true

For n = k

$3 + 33 + 33.... + 333....k \\ = \frac{1}{3} (9+ 99+ 999.....k) \\ = \frac{1}{3} ( ({10}^{1} - 1) + (10 {}^{2} - 1) + ( {10}^{3} - 1) + .....k ) \\ = \frac{1}{3} ((10 + {10}^{2} + ... {10}^{k + 1} ) - (k)) \\ = \frac{1}{3} ( 10 \times \frac{{10}^{k } - 1 } {10- 1} - k ) \\ = \frac{1}{27} ( {10}^{k + 1} - 9k - 10)$

So it is true for n = k

Similar prove for n= k + 1. And it will be proved for n = n