CRE 5 - Miscellaneous | Algebra - Progressions

Answer: Option B

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Explanation :

Let S = 1 + 2x + 3x² + 4x³ + ... (1)

This is an Arithmetic Geometric Progression (AGP). The coefficients are in AP where as the variable is in GP.

Step 1: Multiply the whole equation with x.
⇒ xS = x + 2x² + 3x³ + 4x⁴ + ... (2)

Now, (1) - (2)
⇒ (1 - x)S = 1 + (2x - x) + (3x² - 2x²) + (4x³ - 3x³) + ...
⇒ (1 - x)S = 1 + x + x² + x³ + ...

Now, RHS is an infinite GP whose first term is 1 and common ratio is x.
∴ (1 - x)S = 1/(1 - x)
⇒ S = 1/(1 - x)²

Hence, option (b).

Answer: Option C

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Explanation :

Let S = 1 + 2x + 3x² + 4x³ + ... + nx^n-1    ...(1)

This is an Arithmetic Geometric Progression (AGP). The coefficients are in AP where as the variable is in GP.

Step 1: Multiply the whole equation with x.
⇒ xS = x + 2x² + 3x³ + 4x⁴ + ... + nxⁿ    ...(2)

Now, (1) - (2)
⇒ (1 - x)S = 1 + (2x - x) + (3x² - 2x²) + (4x³ - 3x³) + ... + ((nx^n-1 - (n - 1)x^n-1) - nxⁿ
⇒ (1 - x)S = 1 + x + x² + x³ + ... + x^n-1 - nxⁿ

Now, RHS is a GP (except the last term) whose first term is 1 and common ratio is x.
∴ (1 - x)S = $\frac{{\mathrm{x}}^{\mathrm{n}}-1}{\mathrm{x}-1}$ - nxⁿ

⇒ (x - 1)S =  nxⁿ - $\frac{{\mathrm{x}}^{\mathrm{n}}-1}{\mathrm{x}-1}$

⇒ S =  $\frac{{\mathrm{nx}}^{\mathrm{n}}}{\mathrm{x}-1}$ - $\frac{{\mathrm{x}}^{\mathrm{n}}-1}{{\left(\mathrm{x}-1\right)}^2}$

Hence, option (c).

Answer: Option A

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Explanation :

Let S = 1 + 4x + 9x² + 16x³ + 25x⁴ ... (1)

Step 1: Multiply the whole equation with x.
⇒ xS = x + 4x² + 9x³ + 16x⁴ + 25x⁵ ... (2)

Now, (1) - (2)
⇒ (1 - x)S = 1 + (4x - x) + (9x² - 4x²) + (16x³ - 9x³) +  (25x⁴ - 16x⁴) + ...
⇒ (1 - x)S = 1 + 3x + 5x² + 7x³ + 9x⁴ + ... (3)

Step 2: Again multiply this whole equation with x.
⇒ x(1 - x)S = x + 3x² + 5x³ + 7x⁴ + 9x⁵ ... (4)

Now, (3) - (4)
⇒ (1 - x)²S = 1 + (3x - x) + (5x² - 3x²) + (7x³ - 5x³) +  (9x⁴ - 7x⁴) + ...
⇒ (1 - x)²S = 1 + 2x + 2x² + 2x³ + 2x⁴ + ...

Now, RHS is an infinite GP (except first term) whose first term is 2x and common ratio is x.
∴ (1 - x)²S = 1 + 2x/(1 - x) = (1 + x)/(1 - x)

⇒ S = (1 + x)/(1 - x)​​​​​​​³

Hence, option (a).

Answer: Option C

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Explanation :

Let S =  $\frac{1}{1\times 2}$ + $\frac{1}{2\times 3}$ + $\frac{1}{3\times 4}$ + ...

Now, the first term $\frac{1}{1\times 2}$ = $\frac{2-1}{1\times 2}$ = $\frac{1}{1}$ - $\frac{1}{2}$

Similarly, the n^th term which is $\frac{1}{\mathrm{n}\times (\mathrm{n}-1)}$ can be writtern as $\frac{1}{(\mathrm{n}-1)}$ - $\frac{1}{\mathrm{n}}$

⇒ S = $\left(\frac{1}{1}-\frac{1}{2}\right)$ + $\left(\frac{1}{2}-\frac{1}{3}\right)$ + $\left(\frac{1}{3}-\frac{1}{4}\right)$ .. + $\left(\frac{1}{48}-\frac{1}{49}\right)$ + $\left(\frac{1}{49}-\frac{1}{50}\right)$ + $\left(\frac{1}{50}-\frac{1}{51}\right)$

⇒ S = 1 - $\frac{1}{51}$ = $\frac{50}{51}$

Hence, option (c).

Answer: Option B

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Explanation :

Let S =  $\frac{1}{1\times 3}$ + $\frac{1}{3\times 5}$ + $\frac{1}{5\times 7}$ + ...

∴ S = $\frac{1}{2}\left[\frac{2}{1\times 3}+\frac{2}{3\times 5}+\frac{2}{5\times 7}+...\right]$

Now, the first term $\frac{1}{1\times 3}$ = $\frac{3-1}{1\times 3}$ = $\frac{1}{1}$ - $\frac{1}{3}$

Similarly, the n^th term which is $\frac{2}{(2\mathrm{n}-1)\times (2\mathrm{n}+1)}$ can be writtern as $\frac{1}{(2\mathrm{n}-1)}$ - $\frac{1}{(2\mathrm{n}+1)}$

⇒ S = $\frac{1}{2}\left[\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{99}-\frac{1}{101}\right)\right]$ 

⇒ S = $\frac{1}{2}\left[1-\frac{1}{101}\right]$ = $\frac{50}{101}$

Hence, option (b).

Answer: Option B

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Explanation :

Let S =  $\frac{1}{1\times 2\times 3}$ + $\frac{1}{2\times 3\times 4}$ + $\frac{1}{3\times 4\times 5}$ + ...

Now, the first term $\frac{1}{1\times 2\times 3}$ = $\frac{1}{2}\left(\frac{3-1}{1\times 2\times 3}\right)$ = $\frac{1}{2}\left(\frac{1}{1\times 2}-\frac{1}{2\times 3}\right)$

Similarly, the nth term which is $\frac{1}{\mathrm{n}\times (\mathrm{n}+1)\times (\mathrm{n}+2)}$ can be writtern as $\frac{1}{2}\left(\frac{1}{\mathrm{n}\times (\mathrm{n}+1)}-\frac{1}{(\mathrm{n}+1)\times (\mathrm{n}+2)}\right)$

⇒ S = $\frac{1}{2}\left(\frac{1}{1\times 2}-\frac{1}{2\times 3}\right)$ + $\frac{1}{2}\left(\frac{1}{2\times 3}-\frac{1}{3\times 4}\right)$ + ... + $\frac{1}{2}\left(\frac{1}{49\times 50}-\frac{1}{50\times 51}\right)$ + $\frac{1}{2}\left(\frac{1}{50\times 51}-\frac{1}{51\times 52}\right)$

⇒ S = $\frac{1}{2}\left(\frac{1}{1\times 2}-\frac{1}{51\times 52}\right)$

Hence, option (b).

Answer: Option B

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Explanation :

Let S = $\frac{1}{2^2-1}$ + $\frac{1}{4^2-1}$ + $\frac{1}{6^2-1}$ + ...

a² - b² = (a - b) × (a + b)

∴ S = $\frac{1}{1\times 3}$ + $\frac{1}{3\times 5}$ + $\frac{1}{5\times 7}$ + ...

∴ S = $\frac{1}{2}\left[\frac{2}{1\times 3}+\frac{2}{3\times 5}+\frac{2}{5\times 7}+...\right]$

Now, the first term $\frac{1}{1\times 3}$ = $\frac{3-1}{1\times 3}$ = $\frac{1}{1}$ - $\frac{1}{3}$

Similarly, the n^th term which is $\frac{2}{(\mathrm{n}-2)\times \mathrm{n}}$ can be writtern as $\frac{1}{(\mathrm{n}-1)}$ - $\frac{1}{\mathrm{n}}$

⇒ S = $\frac{1}{2}\left[\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{99}-\frac{1}{101}\right)\right]$ 

⇒ S = $\frac{1}{2}\left[1-\frac{1}{101}\right]$ = $\frac{50}{101}$

Hence, option (b).

Answer: Option B

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Explanation :

T₁ = 1 × 1! = (2 - 1) × 1! = 2 × 1! - 1 × 1! = 2! - 1!

T₂ = 2 × 2! = (3 - 1) × 2! = 3 × 2! - 1 × 2! = 3! - 2!

T₃ = 4! - 3!

... and so on.

∴ Let S = 1 × 1! + 2 × 2! + 3 × 3! + 4 × 4! + ⋯ 10 × 10!

⇒ S = (2! - 1!) + (3! - 2!) + (4! - 3!) + ... + (10! - 9!) + (11! - 10!)

⇒ S = 11! - 1! = 11! - 1

Hence, option (b).

Answer: 1540

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Explanation :

Here, 

T₁ = 1 = 1
T₂ = 3 = 1 + 2
T₃ = 3 = 1 + 2 + 3
... and so on

∴ T_n = 1 + 2 + 3 + ... + n = n(n + 1)/2 = ½ × (n² + n)

⇒ Sum of n terms =  ½ × (Σn² + Σn)

∴ Sum of first 20 terms = ½ × [(1² + 2² + 3² + ... + 20²) + (1 + 2 + 3 + ... + 20)]

= $\frac{1}{2}\left[\frac{20\times 21\times 41}{6}+\frac{20\times 21}{2}\right]$

= ½ × [2870 + 210]

= 1540

Hence, 1540.

Answer: Option C

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Explanation :

Let S = $\frac{1}{2}$ + $\frac{1}{2+4}$ + $\frac{1}{2+4+6}$ + ...

Tn = $\frac{1}{2+4+...+2\mathrm{n}}$ = $\frac{1}{2(1+2+...+\mathrm{n})}$ = $\frac{1}{\mathrm{n}(\mathrm{n}+1)}$ = $\frac{1}{\mathrm{n}}$ - $\frac{1}{\mathrm{n}+1}$

⇒ S = $\left(\frac{1}{1}-\frac{1}{2}\right)$ + $\left(\frac{1}{2}-\frac{1}{3}\right)$ + $\left(\frac{1}{3}-\frac{1}{4}\right)$ + ... + $\left(\frac{1}{9}-\frac{1}{10}\right)$ + $\left(\frac{1}{10}-\frac{1}{11}\right)$

⇒ S = 1 - $\frac{1}{11}$ = $\frac{10}{11}$

Hence, option (c).

Answer: 30

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Explanation :

S_n = n² + n + 1

T_n = S_n - S_n-1

∴ T₁₅ = S₁₅ - S₁₄

⇒ T₁₅ = (15² + 15 + 1) - (14² + 14 + 1)

⇒ T₁₅ = 241 - 211 = 30.

Hence, 30.

Answer: 450

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Explanation :

Given, T_n = n² + n + 1

T₁ = 1² + 1 + 1
T₁ = 2² + 2 + 1
T₁ = 3² + 3 + 1
...
T₁₀ = 10² + 10 + 1

⇒ S₁₀ = (1² + 2² + 3² + ... + 10²) + (1 + 2 + 3 + ... + 10) + (1 + 1 + 1 + ... + 1)

⇒ S₁₀ = 10 × 11 × 21 / 6 + 10 × 11 / 2 + 10

⇒ S₁₀ = 385 + 55 + 10 = 450

Hence, 450.

Answer: 330

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Explanation :

n^th term of this series can be written as n × (n + 1)
⇒ T_n = n² + n

∴ Sum of first n terms of this series = Σn² + Σn

= (1² + 2² + 3² + ... + 9²) + (1 + 2 + 3 + ... + 9)

= $\frac{9\times 10\times 19}{6}$ + $\frac{9\times 10}{2}$

= 285 + 45 = 330

Hence, 330.