Question Number 103130 by chamkunda last updated on 13/Jul/20

  find out the fourth member of the  following formula after expansion  [πx+(2/x)]^8

Commented bychamkunda last updated on 13/Jul/20

please can someone help and answer this  find out the fourth member of the  following formula after expansion  [πx+(2/x)]^8     find out the fourth member of the  following formula after expansion  [πx+(2/x)]^8

Answered by floor(10²Eta[1]) last updated on 13/Jul/20

(a+b)^n =Σ_(k=0) ^n (((n!)/(k!(n−k)!)))a^(n−k) b^k   (πx+(2/x))^8 =Σ_(k=0) ^8 (((8!)/(k!(8−k)!)))(πx)^(8−k) ((2/x))^k   the fourth member is when k=3:  ⇒^(k=3) ((8!)/(3!5!))π^5 2^3 x^2 =448π^5 x^2

Answered by bemath last updated on 13/Jul/20

(πx+(2/x))^8 =Σ_(n=0) ^8 C_n ^8  (πx)^(8−n)  ((2/x))^n   =  (πx)^8 +8(πx)^7 ((2/x))+28(πx)^6 ((2/x))^2 +56(πx)^5 ((2/x))^3 +...

Commented by1549442205 last updated on 13/Jul/20

I agree to this expansion a_3 =((8!)/(2!6!))π^6 x^6 ×(4/x^2 )  =112𝛑^6 x^4 ,excuse me  a_4 =448𝛑^5 x^2  as  above

Commented byfloor(10²Eta[1]) last updated on 13/Jul/20

he wants the fourth member not the third  one

Answered by OlafThorendsen last updated on 13/Jul/20

C_8 ^3 π^5 x^5 (2^3 /x^3 ) = ((8!)/(3!5!))π^5 8x^2  = 448π^5 x^2