Consider `(1 + x + x^(2))^(n) = sum_(r=0)^(n) a_(r) x^(r)` , where ` a_(0), a_(1), a_(2),…, a_(2n)` are
real number and n is positive integer.
The value of ` sum_(r=0)^(n-1) a_(r)` is

Options

A. `(-3^(n) - a_(n))/(2)`
B. `(3^(n) -a_(n))/(2)`
C. `(a_(n) - 3^(n))/(2)`
D. `(3^(n) + a_(n))/(2)`

Answer

b

Explanation

We have , ` (1 + x + x^(2))^(n) = sum_(r=0)^(2n) a_(r) x^(r)` ….(i)
On replacing x by `(1)/(x)` , we get
` (1 + (1)/(x) + (1)/(x^(2)))^(n) = sum_(r=0)^(2n) a_(r) ((1)/(x))^(r)`
` rArr (1 + x + x^(2))^(n) = sum_(r=0)^(2n) a_(r) x^(2n-r)` ...(iii)
From Eqs . (i) and (ii) , we get
` sum_(r=0)^(2n) a_(r) x^(r) = sum_(r=0)^(2n) a_(r) x^(2n-r) `
Equating the coefficient of ` x^(2n - r)` on both sides , we get
` a_(2n-r) = a_(r)`
` 0 le r le 2n ` ...(iii)`
On putting r = 0,1,2,3,...,n-1, n, we get
` a_(2n) = a_(0)`
` a_(2n-1) = a_(1)`
` a_(2n-2) = a_(2)`
`( a_(2n-3) = a_(3) ,)`
`(a_(n+1) = a_(n-1) , a_(n) = a_(n))`
Then , ` a_(0) + a_(1) + a_(2) + ...+ a_(n-1)`
` = a_(n+1) + a_(n+2) + ...+ a_(2n)` ....(iv)
and on putting x = 1 in Eq . (i) , we get
` sum_(r=0) ^(2n) a_(r) = 3^(n)`
` rArr (a_(0) + a_(1) + a_(2) + ...+ a_(n-1) + a_(n) + (a_(n+1) + a_(n+1) + a_(n+2) + ...+ a_(2n)) = 3^(n)`
From Eq . (iv) , we get
` 2(a_(0) + a_(1) + a_(2) + ...+ a_(n-1)) = 3^(n) - a_(n) `
or ` sum_(r=0) ^(n-1) a_(r) = ((3^(n) - a_(n))/(2)`