Question

what is Unit digit of 1+9+9²+9³+.......9¹⁰⁰⁶​

Answer 1

Answer:

1

Step-by-step explanation:

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https://brainly.in/question/3454059

Answer 2

Answer:

Unit digit of Unit digit of 1+9+9²+9³+. . . . . . 9¹⁰⁰⁶​ is 1 .

Step-by-step explanation:

In Context to the given question, we have to find the unit digit of the given progression.

GIVEN EXPRESSION :

⇒ 1+9+9²+9³+. . . . . . 9¹⁰⁰⁶​

SOLUTION:

Let the given progression be T(s)

T(s) = 1 + 9 + 9² + 9³ + 9⁴ + 9⁵ + 9⁶ + . . . . . . + 9¹⁰⁰⁶

BY taking common from every two terms , we get

= 1 + 9(1 + 9) + 9³(1 + 9) + 9⁵( 1 + 9) + 9⁷(1 + 9) +. . . . . . + 9¹⁰⁰⁵( 1 + 9)

= 1 + (1 + 9) [9¹ + 9³ + 9⁵ + 9⁷ + 9⁹ +. . . . . . + 9¹⁰⁰⁵ ]

= 1 + 10 [ 9¹ + 9³ + 9⁵ + 9⁷ + 9⁹ +. . . . . .+ 9¹⁰⁰⁵ ]

Let Sn = 9¹ + 9³ + 9⁵ + 9⁷ + . . . . . . + 9¹⁰⁰⁵

Now, T(s)= 1 + 10 [Sn]

As we know,

{ Any number when multiple with 10 , unit digit of it will be 0 }

∴ unit digit of 10Sn = 0

Now , unit digit of (1 + 10(Sn) )= 1

Therefore, Unit digit of Unit digit of 1+9+9²+9³+. . . . . . 9¹⁰⁰⁶​ is 1 .