The expansion of (1+px+qx^2 )^8 = 1+8x+52x^2 +kx^3 +... What are the values of p ,q and k


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Question Number 148219 by iloveisrael last updated on 26/Jul/21

$The expansion of {(1 + px + {qx}^{2})}^{8}$ $= 1 + 8 x + {52 x}^{2} + {kx}^{3} + . . .$ $What are the values of p, q and k$
	Answered by liberty last updated on 26/Jul/21
	$by Trinomial theorem (1+px+qx^2 )^8 =Σ_(α+β+γ=8) 1^α (px)^β (qx^2 )^γ (i) x−term = ((8!)/(7!1!0!))(px)^1 =8px (ii)x^2 −term=((8!)/(6!2!0!))(px)^2 +((8!)/(7!0!1!))(qx^2 )^1 =(28p^2 +8q)x^2 (iii)x^3 −term=((8!)/(5!3!0!))(px)^3 +((8!)/(6!1!1!))(px)^1 (qx^2 )^1 = (56p^3 +56pq)x^3 comparing coefficients { ((8p=8⇒p=1)),((28p^2 +8q=52⇒q=3)),((56p^3 +56pq=k⇒k=224)) :}$
	$b y T r i n o m i a l t h e o r e m$ ${(1 + p x + {q x}^{2})}^{8} = \sum_{α + β + γ = 8} 1^{α} {(p x)}^{β} {({q x}^{2})}^{γ}$ $(i) x - t e r m = \frac{8!}{7! 1! 0!} {(p x)}^{1} = 8 p x$ $(i i) x^{2} - t e r m = \frac{8!}{6! 2! 0!} {(p x)}^{2} + \frac{8!}{7! 0! 1!} {({q x}^{2})}^{1} = (28 p^{2} + 8 q) x^{2}$ $(i i i) x^{3} - t e r m = \frac{8!}{5! 3! 0!} {(p x)}^{3} + \frac{8!}{6! 1! 1!} {(p x)}^{1} {({q x}^{2})}^{1}$ $= (56 p^{3} + 56 p q) x^{3}$ $c o m p a r i n g c o e f f i c i e n t s$ ${\begin{cases} 8 p = 8 \Rightarrow p = 1 \\ 28 p^{2} + 8 q = 52 \Rightarrow q = 3 \\ 56 p^{3} + 56 p q = k \Rightarrow k = 224 \end{cases}$
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