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If (1 + x – 3x²)¹⁰ = 1 + a₁x + a₂x² + a₂₀x²⁰, then a₂ + a₄ + a₆ + ..+ a₂₀ is equal to

1) [310 + 1] / 2 2) [39 + 1] / 2 3) [310 – 1] / 2 4) [39... View Article

If (1 + x)ⁿ = C₀ + C₁x + C₂x² + … + C_nxⁿ. Then, C₀C₁ + C₁C₂ + …+ C_{n –1} C_n is equal to

1) [2n! / (n – 1)! (n + 1)!] 2) [(2n – 1)! / (n – 1)! (n + 1)!]... View Article

The value of (1 / 81ⁿ) – (10 / 81ⁿ) ²ⁿC₁ + (10² / 81ⁿ) ²ⁿC₂ – (10³ / 81ⁿ) ²ⁿC₃ + ….. + (10²ⁿ / 81ⁿ) is

1) 2 2) 0 3) 1/2 4) 1 Solution: (4) 1 \(\begin{array}{l}\frac{1}{81^{n}}-\frac{10}{81^{n}} 2_{n_{C_{1}}}+\frac{10^{2}}{81^{n}} 2 n_{C_{2}}-\frac{10^{3}}{81^{n}} 2 n_{C_{3}}+\cdots+\frac{10^{2 n}}{81^{n}} \\ =\frac{1}{81^{n}}\left[1-102... View Article

Coefficient of t²⁴ in (1 + t²)¹² (1 + t¹²) (1 + t²⁴) is

1) 12C6 + 3 2) 12C6 + 1 3) 12C6 4) 12C6 + 2 Solution: (4) 12C6 + 2 \(\begin{array}{l}\text... View Article

The value of sum of the series 3 . ⁿC₀ – 8 ⁿC₁ + 13 ⁿC₂ – 18ⁿC₃ + … up to (n + 1) terms is

1) 0 2) 3n 3) 5n 4) None of these Solution: (1) 0 \(\begin{array}{l}3 n_{C_{0}}-8 n_{C_{1}}+13 n_{C_{2}}-18 n_{C_{3}}+. . upto... View Article

(C₀ / 1) + (C₂ / 3) + (C₄ / 5) + (C₆ / 7) + ….. =

1) 2n+1 / (n + 1) 2) (2n+1 – 1) / (n + 1) 3) 2n / (n + 1)... View Article

⁸C₀ / 6 – ⁸C₁ + ⁸C₂ . 6 – ⁸C₃ . 6² +…+ ⁸C₈ . 6⁷ is equal to

1) 0 2) 67 3) 68 4) 58 / 6 Solution: (4) 58 / 6 \(\begin{array}{l}\begin{array}{l} \text { Given }... View Article

The value of ¹⁵C₈ + ¹⁵C₉ – ¹⁵C₆ – ¹⁵C₇ is

1) – 1 2) 0 3) 1 4) None of these Solution: (2) 0 \(\begin{array}{l}15 c_{c_{8}}+15 c_{9}-15_{c_{6}}-15_{C_{7}} \\ =15_{C_{7}}+15_{C_{6}}-15_{c_{6}}-15_{C_{7}} \\... View Article

¹⁵C₀ . ⁵C₅ + ¹⁵C₁ . ⁵C₄ + ¹⁵C₂ . ⁵C₃ + ¹⁵C₃ . ⁵C₂ + ¹⁵C₄ . ⁵C₁ is equal to

1) 220 – 25 2) 20! / 5! 15! 3) (20!/5! 15!) – 1 4) [20! / (5! 15!)] –... View Article

^{n – 2}Cr + 2^{n -2}C_{r -1} + ^{n -2}C_{r – 2} is equal to

1) n+1Cr 2) nCr 3) n+1Cr+1 4) n -1Cr Solution: (2) nCr \(\begin{array}{l}n-2_{c_{r}}+2 n-2_{c_{r-1}}+n-2_{c_{r-2}} \\ =n-2_{c_{r}}+n-2_{c_{r-1}}+n-2_{c_{r-1}}+n-2_{c_{r-2}} \\ =\left.n-1_{c_{r}}+n-1_{c_{r-1}} \text {... View Article

If the sum of the coefficients in the expansion of (a²x² – 6ax + 11)¹⁰, where a is constant is 1024, then the value of a is

1) 5 2) 1 3) 2 4) 3 5) 4 Solution: (4) 3 \(\begin{array}{l}\text \ Given, \ sum \ of... View Article

Let (1 + x)ⁿ = 1 + a₁x + a₂x² + …+ a_nxⁿ. If a₁, a₂ and a₃ are in AP, then the value of n is

1) 4 2) 5 3) 6 4) 7 5) 8 Solution: (4) 7 \(\begin{array}{l}(1+x)^{n}=n_{c_{0}}+n_{c_{1}} x+n_{c_{2}} x^{2}+\cdots+n_{c_{n}} x^{n}\\ (1+x)^{n}=1+a_{1} x+a_{2} x^{2}+\cdots+a_{n}... View Article

For r = 0, 1, …, 10 let A_r, B_r and C_r denote, respectively the coefficient of x^r in the expansion of (1 + x)¹⁰, (1 + x)²⁰ and (1 + x)³⁰. Then ∑_r=1¹⁰ A_r (B₁₀ B_r – C₁₀ A_r) =

1) B10 – C10 2) A10 (B102 – C10A10) 3) 0 4) C10 – B10 Solution: (4) C10 – B10... View Article

The expression ⁿC₀ + 2ⁿC₁ + 3ⁿC₂ + (n + 1)ⁿC_n is equal to

1) (n + 1)2n 2) 2n(n + 2) 3) (n + 2)2n-1 4) (n + 2)2n+1 Solution: (3) (n +... View Article

If C₀, C₁, … C_n denote the binomial coefficients in the expansion of (1 + x)ⁿ. Then, the value of C₁ – ²C₂ + ³C₃ – ⁴C₄ + ….(up to n terms) is

1) 2n 2) 2-n 3) 0 4) 1 Solution: (3) 0 \(\begin{array}{l}(1+x)^{n}=C_{0}+C_{1} x+C_{2} x^{2}+C_{3} x^{3}+\cdots+C_{n} x^{n}\\ \text \ Differentiating \... View Article

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