If (1+b y)n=(1+8 y+24 y2+ ldots.) where n ∈ N then the value of b and n are respectively-

Question

If (1+b y)n=(1+8 y+24 y2+ ldots.) where n ∈ N then the value of b and n are respectively- (A) 4,2 (B) 2,-4 (C) 2,4  (D) -2,4

Tardigrade Admin · Accepted Answer

(1+b y)n=1+8 y+24 y2+ (1+b y)n=1+n ⋅ b y+(n(n-1)/2) ⋅ b2 y2+ ldots ldots ldots by comparison bn =8.....(1) (n(n-1)/2) b2=24.....(2) 4 n-4=3 n n =4 and b=2

Q. If $(1 + b y)^{n} = (1 + 8 y + 24 y^{2} + \dots .)$ where $n \in N$ then the value of $b$ and $n$ are respectively-

4,2

$2, - 4$

$2, 4$

$- 2, 4$

$(1 + b y)^{n} = 1 + 8 y + 24 y^{2} +$
$(1 + b y)^{n} = 1 + n \cdot b y + \frac{n ( n - 1 )}{2} \cdot b^{2} y^{2} + \dots\dots\dots$
by comparison
$bn = 8.....$ (1)
$\frac{n ( n - 1 )}{2} b^{2} = 24.....$ (2)
$4 n - 4 = 3 n$
$n = 4$
and $b = 2$

Q. If (1+by)n=(1+8y+24y2+….) where n∈N then the value of b and n are respectively-

4,2

2,−4

2,4

−2,4

(1+by)n=1+8y+24y2+ (1+by)n=1+n⋅by+2n(n−1)​⋅b2y2+……… by comparison bn=8.....(1) 2n(n−1)​b2=24.....(2) 4n−4=3n n=4 and b=2

Q. If $(1 + b y)^{n} = (1 + 8 y + 24 y^{2} + \dots .)$ where $n \in N$ then the value of $b$ and $n$ are respectively-

$2, - 4$

$2, 4$

$- 2, 4$

$(1 + b y)^{n} = 1 + 8 y + 24 y^{2} +$
$(1 + b y)^{n} = 1 + n \cdot b y + \frac{n ( n - 1 )}{2} \cdot b^{2} y^{2} + \dots\dots\dots$
by comparison
$bn = 8.....$ (1)
$\frac{n ( n - 1 )}{2} b^{2} = 24.....$ (2)
$4 n - 4 = 3 n$
$n = 4$
and $b = 2$