Question

HELP ----> [Binomial Theorem]

Answer 1

Find the positive value of "a" so that the co-efficient of x^5 is equal to that of x^15 in the expansion
\[\huge (x^{2} + \frac{ a }{ x^{3} })^{10}\]

Answer 2

http://prntscr.com/5e5j12

Answer 3

the coefficient for x^5 i am getting is
\[\huge C _{5}^{10}a^{5}\]

Answer 4

http://prntscr.com/5e5jep

Answer 5

expanding it is useless actually

Answer 6

:D ikr but since woli can do it i dont mind :P

Answer 7

Hint: the seciond term is 10 (x*2)^9 * a/x^3

Answer 8

the coeffiecient i am getting for x^15 is
\[\huge C _{1}^{10}a^1\]

Answer 9

right

Answer 10

i equal them right

Answer 11

yes

Answer 12

252a^5 = 10 a --- > right ?

Answer 13

\[\huge a^4 = \frac{ 10 }{ 252 }\]
\[\huge a= \sqrt[4]{\frac{ 10 }{ ? }}\]

Answer 14

\[\huge \huge a= \sqrt[4]{\frac{ 10 }{ 252 }}\]

Answer 15

the ? is 252

Answer 16

sorry i have to go right now

Answer 17

answer is given different though

Answer 18

@ganeshie8 help

Answer 19

120 x^3= 10 a
120 x^3- 10 a=0
10a(12a^2-1)=0
a=0
or
a=1/12
:-|