Coefficient In A Septic Expansion

By binomial theorem it is $2^4 \cdot \binom{7}{3} = \boxed{560}$ .

By the Binomial Theorem, $\left(x+2\right)^{7}=\sum_{i=0}^{7} \dbinom {7}{i} x^{i}2^{7-i}$ .

Therefore, to get what happens when the power of x is 3, the term would be $\dbinom {7}{3} x^{3}2^{4}=35 \times x^{3} \times 16=560x^{3}$ .

Therefore, the coefficient of $x^{3}$ in the expansion of $\left(x+2\right)^{7}$ is $\boxed {560}$ .

The given expression is $(x+2)^{7}$ . Let $T_{r+1}$ be the $(r+1)$ th term of the expansion. Since the expansion starts with the first term with $x^{7}$ and with each successive term, the power of $x$ decreases, so the $5$ th term is the required term with $x^{3}$ . We shall use the identity $T_{r+1}=nCr\times x^{n-r} a^{r}$ for the expansion of $(x+a)^{n}$ . In this case, we have $a=2$ and $n=7$ .

So, $T_{5}=T_{4+1}=7C4\times x^{7-4} 2^{4} = \frac{7!}{4!\times (7-4)!}\times x^{3} \times 16 = \frac{7\times 6\times 5\times 4!}{4! \times 6}\times x^{3} \times 16 = 35\times 16 x^{3} = \boxed{560x^{3}}$

So, the coefficient of the term $=\boxed{560}$

use binomial coefficients we get $(x+2)^7=\sum_{k=0}^{7}\begin{pmatrix}7 \\ k \end{pmatrix}x^k 2^{n-k}$ so the coefficient of x^3 is $\binom{7}{3}2^{4}=560$

(x+2)^7=x^7+7C1(x^6)(2)+7C2(x^5)(2)^2+7C3(x^4)(2)^3+7C4(x^3)(2)^4+7C5(x^2)(2)^5+7C6(x^1)(2)^6+2^7 From 7C4(x^3)(2)^4=x^3(35*16)=560x^3

we can use Binomial Newton to solve this. if $y$ is coefficient of $x^{3}$ at $(x+2)^{7}$

$y$ = $7C4.x^{3}.2^{4}$

$y$ = $35.x^{3}.16$

$y$ = $560x^{3}$

(7/3) a^3 b^4

35 x^3 2^4

560*x^3

binomial expansion... x^3 has coefficient 7C3 X 2^4. Which is equal to 560

7c4*16=560

Tr+1=nCr (x)^(n-r) y^r (x+2)^7 in this problem x=1,y=2,n=7. so i can fill the value in above equation. Tr+1=7C4 (1)^(7-r) 2^r here we need r value so 7-r=3, r=7-3, finally r=4. T4+1=7C4 (1)^(7-4) 2^4 T5=7C4 (1) 16 here 7C4=35 ,,( nCr=n!/r!(n-r)!) T5=35 1 16 T5=560

Coefficient of x^3 = 7C4 * x^3*2^4, and 7 Combination 4 = 35 and 2 ^4 =16 Therefore, 35 x 16 = 560

7 C 4 = 7!/3!4! = 35

am i the only one who use pascal triangle to solve this?

560

applying Tr+1=nCr ((x)^n-r) y^r x=x,y=2,n=7,r=4 T4+1=7C4 ((x)^7-4) 2^4 T5=35 x^3 16 coefficient x^3=560

7C3*2^4= 560

7C4 x 2^4 = 35 X 16 = 560

One of the way of solving this is using BINOMIAL EXPANSION. (x+y)^n= nC0 x^n y^0 + nC1 x^(n-1) y^1 + nC2 x^(n-2) y^2 + ... + nCn x^0 y^n. Evidently, the coefficient of x^3 should be (nC4 y^4). Here, n=7 & y=2. Thus, coeff. of x^3 = 7C4 2^4 = 560.

7c4 x^(7-4) 2^4 .... SO the coefficient of x^3 is 7c4*2^4 :)

using binomial theory,,(x+y)^n then put n=7 and y=2

7C3 x 2^(7-3) x X^3 = 560X^3

Coefficient of x^3 in (x+2)^7 is 7C3 x 2^(7-4)=560

By the binomial theorem, the term with x^{3} should be

C(7,3) x^{3} \times 2^{7-4}.

Hence the coefficient of x^{3} is C(7,3)*2^{4}

C^{7}r= x^7-r * 2^r (formula) Now for the power of x to be 3, x^7-r should be =3. Therefore r=4 (7-r=3) now the coefficient would be c^7 4 * 2^4, i,e 7!/3! 4! (nCr= n!/r! *(n-r)!) Therefore coefficient is 35 16=560

To finding rth power coefficient we know=nCr-1 x^{n-r+1} a^{r-1}

so coefficient= 7C4 * x^{7-4} * 2^{4}

=560

ans =560

7c4 . (x)^3 (2)^4 = 35 . 16 . x^3 = 560 x^3

Pascal's triangle :)

C(7,3).x^{3}.2^{4}

the term in Binomial expression (x + y)^n is nCr x^(n – r ) y^r

in (x + 2)^7 to get x^3 term (n – r) = 3 here n = 7 and r = 4

so term is 7C4 x^3 y^4

= 7!/4!×3! x^3 (2)^4 = 35 × 16 = 560

in the expansion if (x+y)^n ------> Tr+1 = nCr * y^r * x^n-r. so in (x+2)^7 ------> Tr+1 = 7Cr * 2^r * x^7-r ------> we want x^3 , so 7-r = 3 ------>. r = 4. ------> T5 = 7C4 * 2*4 * x^3 ------> then the coefficient of x^3 = 7C4 * 2^4 = 560