What is C(n,1)+C(n,2)++C(n,n) equal to [NDA-I 2021]

Correct Answer is : 1+2+22+23+4+2n−11+2+2²+2³+₄+2ⁿ-1}1+2+22+23+4+2n−1

Correct Answer is : 1+2+22+23+4+2n−11+2+2²+2³+₄+2ⁿ-1}1+2+22+23+4+2n−1

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NDA Math Binomial Theorem

Question : 18 of 50

Marks: +1, -0

What is

C(n,1)+C(n,2)+\dots+C(n,n)

Solution: 👈: Video Solution

Concept

(1+x)^{n}={}^{n}C_{0}\times 1^{n-0}\times x^{0}+{}^{n}C_{1}\times 1^{n-1}\times x^{1}+{}^{n}C_{2}\times 1^{n-2}\times x^{2}+\dots+{}^{n}C_{n}\times 1^{n-n}\times x^{n}

n^{\;\text{th }\;}

term of the G.P. is an

= ar^{n-1}

Sum of

n

terms

= s = \;\frac{a(r^{n}-1)}{r-1}\;\frac{a(r^{n}-1)}{r-1};

where

r > 1

Calculation

Sum of

n

terms

= s = \;\frac{a(1-r^{n})}{1-r}\;\frac{a(1-r^{n})}{1-r};

where

r < 1

C(n,1)+C(n,2)+\dots+C(n,n)

= {}^nC_1+{}^nC_2+\dots+{}^nC_n

= {}^nC_0+{}^nC_1+{}^nC_2+\dots+{}^nC_n-{}^nC_0

= (1+1)^n-{}^nC_0

= 2^n-1 = \frac{2^n-1}{2-1} \frac{2^n-1}{2-1} = 1 \times \frac{2^n-1}{2-1} \frac{2^n-1}{2-1}

Comparing it with a G.P sum

= a \times \frac{r^n-1}{r-1} \frac{r^n-1}{r-1}

, we get

a=1

and

r=2

So,

2^n-1=1+2+2^2+\dots+2^{n-1}

which will give us

n

terms in total.

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