Question

if Ts=1+9+9^2+9^3+9^4......+9^100 and unit digit of Ts is n then find n+9

Answer 1

Other Method :
Final Answer : 1

T(s) = 1+ 9 + 9^2 + ........9^100.

Steps:
1) Factorise the expression.
2) Units digit of 10 is 0 .

Now,
$= > 1 +9 + {9}^{2} + {9}^{3} +..... {9}^{100} \\ = > 1 + 9(1 + 9) + {9}^{3} (1 + 9) + \\ ..... {9}^{99} (1 + 9) \\ = > 1 + (1 + 9)(9 + {9}^{3} + ..... {9}^{99} ) \\ = > 1 + 10(9 + {9}^{3} + {9}^{5} + .... {9}^{99} ) \\ = > 1 + multiple \: of \: 10.$

Now,
Units digit of T(s):
= units digit of (1+ multiple of 10)
= units digit of (1+0)
= 1 .

Therefore, n= 1.

And (n+9) = 1+9=10.

Answer 2

Jinkazama1 method is superb , If possible give him brainliest :)

I want to mention same method
Ts = 1 + 9 + 9² + 9³ + 9⁴ + 9⁵ + 9⁶ + ...... + 9¹⁰⁰
= 1 + 9(1 + 9) + 9³(1 + 9) + 9⁵( 1 + 9) + 9⁷(1 + 9) + ........ + 9⁹⁹( 1 + 9)
= 1 + (1 + 9) [9¹ + 9³ + 9⁵ + 9⁷ + 9⁹ + ........ + 9⁹⁹ ]
= 1 + 10 [ 9¹ + 9³ + 9⁵ + 9⁷ + 9⁹ + ....... + 9⁹⁹ ]

Let Sn = 9¹ + 9³ + 9⁵ + 9⁷ + ..... + 9⁹⁹
Now, Ts = 1 + 10Sn
As you know, any number when multiple with 10 , unit digit of it will be 0
∴ unit digit of 10Sn = 0
Now , unit digit of (1 + 10Sn) = 1

A/C to question,
Unit digit of Ts = n
So, n = 1

∴ n + 9 = 1 + 9 = 10