$\begingroup$ I was thinking of doing $\binom{8}{4}$ but not sure if right.
asked Sep 29, 2015 at 5:27
$\endgroup$ 8 $\begingroup$ Pre-expansion, there are $8$ factors of $2x - y + 5$. From those $8$ factors, choose the $3$ that contribute to the $x^3$, from the remaining $5$ factors, choose the $4$ that will contribute to $y^4$. There is only 1 factor left so it chooses itself. Post-expansion, that's ${8 \choose 3}{5 \choose 4}{1 \choose 1}$ terms with $x^3y^4$, and taking into account the initial coefficients of $\{2, -1, 5\}$, the final coefficient is: $${8 \choose 3}{5 \choose 4}2^3(-1)^45^1 = 11200$$ answered Sep 29, 2015 at 6:34
DanielVDanielV 22.4k5 gold badges36 silver badges68 bronze badges $\endgroup$ 1 $\begingroup$ One has $$(2x-y+5)^8 = \sum_{i=0}^8 {8 \choose i} (2x-y)^i 5^{8-i} = \sum_{i=0}^8 {8 \choose i} 5^{8-i}\sum_{j=0}^i {i\choose j} (2x)^jy^{i-j}(-1)^{i-j}.$$ Then, one gets the coefficient of $x^3y^4$ is $${8 \choose 7}5^{8-7}{7\choose 3}2^3.$$ answered Sep 29, 2015 at 5:49
GAVDGAVD 7,1401 gold badge15 silver badges30 bronze badges $\endgroup$ 2 $\begingroup$ Notice, the binomial expansion $$(2x-y+5)^8=^{8}C_0(2x-y)^85^0+\color{red}{^{8}C_1(2x-y)^75^1}+^{8}C_2(2x-y)^65^2+^{8}C_3(2x-y)^55^3+^{8}C_4(2x-y)^45^4+\dots +^{8}C_8(2x-y)^05^8$$ In the given term $x^3y^4$, the sum of powers $3+4=7$ hence in the above expansion there is only one term $^{8}C_1(2x-y)^75^1$ which has sum of powers $7$ hence using binomial expansion of $^{8}C_1(2x-y)^75^1$ as follows $$^{8}C_1(2x-y)^75^1=\color{blue}{5(^{8}C_1)}[^{7}C_0(2x)^7(-y)^0+^{7}C_1(2x)^6(-y)^1+^{7}C_2(2x)^5(-y)^2+^{7}C_3(2x)^4(-y)^3+\color{red}{^{7}C_4(2x)^3(-y)^4}\ldots +^{7}C_7(2x)^0(-y)^7]$$ Hence, the coefficient of $x^3y^4$ of $5(^{8}C_1)(^{7}C_4)(2x)^3(-y)^4$ is given as $$=5(^{8}C_1)(^{7}C_4)(2)^3(-1)^4$$ $$=5(8)(35)(8)(1)=\color{red}{11200}$$ answered Sep 29, 2015 at 6:29
$\endgroup$ $\begingroup$ HINT: The $r,0\le r\le8$th term of $$(2x-y+5)^8=\{(2x+5)-y\}^8$$ is $$\binom8ry^{8-r}(2x+5)^{8-r}(-y)^r$$ So, we need $r=4$ So, we focus on $$\binom84y^{8-4}(2x+5)^{8-4}(-y)^4$$ Now the $n,0\le n\le4$th term of $$(2x+5)^4$$ is $$\binom4n(2x)^{4-n}5^n$$ We need $4-n=3$ answered Sep 29, 2015 at 5:44
lab bhattacharjeelab bhattacharjee 269k17 gold badges199 silver badges313 bronze badges $\endgroup$ $\begingroup$ You can also solve it using the multinomial theorem.
daOnlyBG 2,6477 gold badges21 silver badges36 bronze badges answered Sep 29, 2015 at 13:50
$\endgroup$ Coefficient of#x^8 y^5# is#1287# Explanation:We know #(a+b)^n= nC_0 a^n*b^0 +nC_1 a^(n-1)*b^1 + nC_2 a^(n-2)*b^2+..........+nC_n a^(n-n)*b^n# Here #a=x,b=y,n=13# We know, #n C_r = (n!)/(r!*(n-r)!# #x^8 *y^5# will be in #6# th term #T_6= 13 C_5 * x^(13-5)* y^5= 13 C_5 x^8*y^5# #13 C_5 =(13!)/(5!*(13-5)!) =1287# Coefficient of #x^8 y^5# is #1287# [Ans] |