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

The expansion of (1+px+qx^2 )^8 = 1+8x+52x^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)) :}