Mathematics Test 6

Result

Mathematics Test 6

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Weekly Quiz Competition

Question 1
4 / -1

Let n be an odd natural number greater than 1. Then the number of zeros at the end of the sum 99ⁿ + 1 is-

A
3

B
4

C
2

D
None of these

Solution

99ⁿ + 1 = (100 – 1)ⁿ + 1 = ⁿC₀ (100)ⁿ (–1)⁰ + ⁿC₁ (100)^{n – 1} (–1)¹+ … + ⁿC_{n – 1} (100) (–1)^{n –1} + ⁿC_n (100)⁰ (–1)ⁿ + 1

= 100 m + n × 100 – 1 + 1

= 100 m + n × 100 ; m ∈ I⁺

n odd integer greater than 1.

Hence, at the end of sum, at least two zeros will be.

∴Option C is correct Answer.
Question 2
4 / -1

If C₀, C₁, C₂, ............ C_n are the Binomial coefficients in the expansion of (1 + x)ⁿ, n being even, then C₀ + (C₀ + C₁) + (C₀ + C₁ + C₂) + ..... + (C₀ + C₁ + C₂ + ........+ C_n–1) is equal to

A
n . 2ⁿ

B
n.2^n–1

C
n.2ⁿ⁺³

D
n.2^n–3

Solution

(1 + x)ⁿ = ⁿC₀ + ⁿC₁.x + ⁿC₂ x² + ⁿC₃ x³ + ⁿC₄ x⁴ + ⁿC₅ x⁵ + … + ⁿC_n xⁿ

Differentiating w.r.t. x we get

n(1 + x)^{n – 1} = 0 + ⁿC₁+ 2x ⁿC₂ + 3x²ⁿC₃ + 4x³ⁿC₄ + 5x⁴ⁿC₅ + … + nx^{n – 1 n}C_n

Put x = 1, we get

n.(2)^{n – 1} = ⁿC₁ + 2ⁿC₂ + 3 ⁿC₃ + 4ⁿC₄ + 5ⁿC₅ + … … + nⁿC_n

Since ⁿC₁ = ⁿC_{n – 1}& ⁿC₀ = ⁿC_n
ⁿC₂ = ⁿC_{n – 2}

n.2^{n – 1} = C₀ + (C₀ + C₁) + (C₀ + C₁ + C₂) + (C₀ + C₁ + C₂ + C₃) + ... + (C₀ + C₁ + C₂ + C₃ + …+C_{n – 1})

Hence, option B is correct Answer.