Subham's coefficient sum

The series forms PASCAL's traingle

                     1

1, 1

1, 2, 1

1, 3, 3, 1

1, 4, 6, 4, 1

1, 5, 10, 10, 5 , 1

Therefore, sum of 1+5+10+10+5+1 = 32

Well, according to the binomial theorem, the sum of the coefficients is

$\sum_{k=0}^{5} \binom{5}{k} = 32$

It's a Newton's Binomial, which can be write how: $(1+x)^5)$ = $\sum_{k=1}^5*x^{n-k}*y^k*(\frac{n}{k})$ , where $n$ = exponent, $a$ = a, $b$ = b. So, $1$ + $5$ + $10$ + $10$ + $5$ + $1$ = $\boxed{32}$

Use Pascal triangle .Since we have 5 in the exponent,we get 6 terms. So we have to use 6th row of Pascal triangle. Since we have 1 and x ,we can directly add the numbers in the 6th row to get the sum of the coefficients. Sixth row of Pascal triangle is 1 5 10 10 5 1 Their sum =1+5+10+10+5+1 = 32

Let $P(x)=(1+x)^5$ , so $P(1)$ is the sum of the coefficients of $P(x)$ .

this is simple problem but so hard to find its simple. it use pascal triangle just look at the number of expansion, it's 5. so, 2^5 = 32

5C1+5C2+5C3+5C4+5C5 =1+5+10+10+5+1 =32 topic: Binomial equation, Combination & permutation.

just make use of pascal triangle so the coefficients will be 1 5 10 10 5 1 so 1+5+10+10+5+1=32

(1+1)^5=32

(1+x)^5 = 5C0. X^(5-0) + 5C1. X^(5-1) + 5C2. X^(5-2) + 5C3. X^(5-3) + 5C4. X^(5-4) + 5C5. X^(5-5) = 1x^5 + 5.x^4 + 10x^3 + 10x^2 + 5x + 1. So, sum of all coefficients of (1+x)^5 = 1 + 5 + 10 + 10 + 5 + 1 = 32. Answer : 32

for a polynomial, (1+x)^n, the rth co-efficient is rCn........in current case r wil range from 1 to 5 and n = 5....co-efficients are 1, 5, 10, 10, 5, 1 ( 6 terms).........thus sum is 32

5c0+5c1+5c2+5c3+5c4+5c5 = 32

(1+x)^5=1+5+10+10+5+1

We use here the binomial theorem to find the expansion of the expression $(1+x)^5$ .Also, we must know that the value of the binomial coefficient $\binom{n}{r}=\frac{n!}{r!\times (n-r)!}$ . From the binomial theorem, we know that ---

$(1+x)^5=\binom{5}{0}+\binom{5}{1}x+\binom{5}{2}x^2+\binom{5}{3}x^3+\binom{5}{4}x^4+\binom{5}{5}x^5$

Now, Sum of the coefficients $=\binom{5}{0}+\binom{5}{1}+\binom{5}{2}+\binom{5}{3}+\binom{5}{4}+\binom{5}{5}$

$=1+5+10+10+5+1 = \boxed{32}$

Put x=1 in expansion of (1+x)^n.... You will find that sum of co-efficients is 2^n... Here as n=5, sum of co-efficients is 2^5=32.....

formula is 2^{n} 2^{5}=32

sum C(5,k) , k=0 to 5. Newton Binomial

1, 1 (by pascals triangle)

1, 2, 1

1, 3, 3, 1

1, 4, 6, 4, 1

1, 5, 10, 10, 5 , 1

Therefore, sum of 1+5+10+10+5+1 = 32

by binomial expansion as well

1+5+10+10+5+1 = 32

expansion of (1+x)^5=1+ 5x + 10x^2 + 10x^3 + 5x^4 + x^5 now add every coefficients i.e. 1+5+10+10+5+1