Question

Let [x] denote greatest integer less than or equal to x. If for n ∈ N, (1 – x + x³)ⁿ = `sum_("j" = 0)^(3"n")"a"_"j"x^"j"`, then `sum_("j" = 0)^([(3"n")/2]) "a"_(2"j") + 4sum_("j" = 0)^([(3"n" - 1)/2])"a"_(2"j" + 1)` is equal to ______.

पर्याय

• 1

• n

• 2^n–1

• 2

Answer 1

Let [x] denote greatest integer less than or equal to x. If for n∈N, (1 – x + x³)ⁿ = `sum_("j" = 0)^(3"n")"a"_"j"x^"j"`, then `sum_("j" = 0)^([(3"n")/2]) "a"_(2"j") + 4sum_("j" = 0)^([(3"n" - 1)/2])"a"_(2"j" + 1)` is equal to 1.

Explanation:

Given that (1 – x + x³)ⁿ = `sum_("j" = 0)^(3"n")"a"_"j"x^"J"`

⇒ (1 – x + x³)ⁿ = `"a"_0 + "a"_1x + "a"_2x^2 + "a"_3x^3 + ....... + "a"_(3"n")x^(3"n")`  ...(i)

We have to find the value of `sum_("j" = 0)^([(3"n")/2]) "a"_(2"j") + 4sum_("j" = 0)^([(3"n" - 1)/2])"a"_(2"j" + 1)`

Here, `sum_("j" = 0)^([(3"n")/2]) "a"_(2"j")` = a₀ + a₂ + a₄ + ..... and `sum_("j" = 0)^([(3"n" - 1)/2])"a"_(2"j" + 1)` = a₁ + a₃ + a₅ + .....

Put x = 1 in equation (i)

1 = a₀ + a₁ + a₂ + a + ... + a_3n  ...(ii)

Put x = –1 in equation (i)

1 = a₀ – a₁ + a₂ – a₃ + ......(–1)³ⁿa³ⁿ  ...(iii)

After adding (ii) and (iii) we get

2 = 2(a₀ + a₂ + a₄ + ......)

⇒ a₀ + a₂ + a₄ + ..... = 1  

i.e. `sum_("j" = 0)^([(3"n")/2]) "a"_(2"j")` = 1   ...(iv)

and a₁ + a₃ + a₅ + .... = 0

i.e. `4sum_("j" = 0)^([(3"n" - 1)/2])"a"_(2"j" + 1)` = 0  ...(v)

Add equation (iv) and (v)

⇒ `sum_("j" = 0)^([(3"n")/2]) "a"_(2"j" + 1) + 4sum_("j" = 0)^([(3"n" - 1)/2])"a"_(2"j" + 1)` = 1