Question

nC1 + nC2+nC3+nC4 + .....+nCn equals
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Answer 1

Step-by-step explanation:

In the binomial expansion

(a+b)^n = nC0a^n. b^0 + nC1 a^(n-1)b^1 + nC2a^(n-2)b^2 + etc

If you now let a=b=1

(1+1)^n =2^n = nC0 + nC1 +nC2 +. etc .

In fact what we have here is the number of subsets of size r taken from a set of n objects.

nC0 = 1 = number of sets containing zero elements

nC1 = n = " ". " " 1. "

nC2 = (n)(n-1)/2 ". ". ". ". 2 ". etc

Total = nC0 + nC1+ nC2 +. = 2^n =total number of subsets in a set of n elements.

Again if a=1 and b=-1

(1-1)^n =0 = nC0 - nC1 + nC2 - nC3

nC0 +n C2 + nC4 + = nC1 + nC3 + nC3 +.

= (1/2) sigma nCr = (1/2) .2^n = 2^(n-1)

Answer 2

nC1 + nC2+nC3+nC4 + ..... +nCn = $2^{n} -1$

Given:

nC1 + nC2+nC3+nC4 + .....+nCn

To Find:

Value of nC1 + nC2+nC3+nC4 + .....+nCn

Solution:

WKT;

$(1+x)^{n}$ = $nC_{0} x^{0} + nC_{1} x^{1} +.... nC_{n} x^{n}$

$nC_{r}$ =  n!/r!(n-r)!

Putting x = 1;

=> $(1+1)^{n}$ = $nC_{0} 1^{0} + nC_{1} 1^{1} +.... nC_{n} 1^{n}$

=> $2^{n}$ = 1 + nC1 + nC2 + nC3 + … + nCn

=> nC1 + nC2 + nC3 + … + nCn = $2^{n}$ -1

Hence, the sum of the series nC1 + nC2 + nC3 + … + nCn equals $2^{n}$ -1.

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