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Evaluate : 9 + 99 + 999 + ........... up...

Evaluate : 9 + 99 + 999 + ........... upto n terms.

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To evaluate the expression \( 9 + 99 + 999 + \ldots \) up to \( n \) terms, we can follow these steps: ### Step 1: Rewrite the terms We can express each term in a more manageable form: - The first term is \( 9 = 10^1 - 1 \) - The second term is \( 99 = 10^2 - 1 \) - The third term is \( 999 = 10^3 - 1 \) - Continuing this pattern, the \( k \)-th term can be expressed as \( 10^k - 1 \). Thus, we can rewrite the entire sum as: \[ 9 + 99 + 999 + \ldots + \text{(up to n terms)} = (10^1 - 1) + (10^2 - 1) + (10^3 - 1) + \ldots + (10^n - 1) \] ### Step 2: Combine the terms Now, we can combine the terms: \[ = (10^1 + 10^2 + 10^3 + \ldots + 10^n) - n \] Here, \( n \) is subtracted because we have \( n \) terms of \(-1\). ### Step 3: Sum the geometric series The sum \( 10^1 + 10^2 + 10^3 + \ldots + 10^n \) is a geometric series where: - The first term \( a = 10 \) - The common ratio \( r = 10 \) - The number of terms \( n \) The formula for the sum of the first \( n \) terms of a geometric series is: \[ S_n = a \frac{r^n - 1}{r - 1} \] Substituting the values: \[ S_n = 10 \frac{10^n - 1}{10 - 1} = 10 \frac{10^n - 1}{9} \] ### Step 4: Substitute back into the equation Now we substitute this back into our expression: \[ = \left(10 \frac{10^n - 1}{9}\right) - n \] ### Step 5: Final expression Thus, the final expression for the sum \( 9 + 99 + 999 + \ldots \) up to \( n \) terms is: \[ = \frac{10(10^n - 1)}{9} - n \]
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ICSE-MIXED PRACTICE -SET B
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  2. If a, b, c are in G.P. and also a, b, c are in A.P. Prove that: 1/a+...

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  3. Evaluate : 9 + 99 + 999 + ........... upto n terms.

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  4. Find the point on the y-axis whose distances from the points (3, 2) an...

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  6. A straight line makes on the co-ordinate axes positive intercepts whos...

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  7. The line 3x - 4y + 12 = 0 meets x-axis at point A and y-axis at point ...

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  8. The line 3x - 4y + 12 = 0 meets x-axis at point A and y-axis at point ...

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  9. In a triangle PQR, L and M are two points on the base QR, such that an...

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  10. In a triangle PQR, L and M are two points on the base QR, such that an...

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  11. In a triangle PQR, L and M are two points on the base QR, such that an...

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  12. In a rectangle ABCD, its diagonal AC = 15 cm and angleACD = alpha If ...

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  13. The given figure shows, AB is a diameter of the circle. Chords AC and ...

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  14. Use ruler and compasses for this question. Construct an isosceles tr...

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  15. Use ruler and compasses for this question. Draw AD, the perpendicula...

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  16. Use ruler and compasses for this question. Draw a circle with centre...

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  17. In triangle ABC, angleBAC = 90^@, AB = 6 cm and BC = 10 cm. A circle ...

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  18. A conical vessel of radius 6 cm and height 8 cm is completely filled w...

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  19. Prove that : (1+cotA)/(cosA)+(1+tanA)/(sinA)=2(secA+"cosec"A)

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  20. Prove that : sqrt((1+sinA)/(1-sinA))-sqrt((1-sinA)/(1+sinA))=2tanA

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