What is {n}{1} + {n}{2} + ·s + {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

1 + 2 + 2² + 2³ + \;\; + 2ⁿ

1 + 2 + 2² + 2³ + ₄ + 2ⁿ-1}

2 + 2² + 2³ + _{-----} + 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

Test Index

UPSC NDA I 2021 Mathematics Solved Paper

Question : 1 of 120

Marks: +1, -0

What is

\binom{n}{1} + \binom{n}{2} + \cdots + \binom{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

\binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n}

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

= {}^{n}C_{0} + {}^{n}C_{1} + {}^{n}C_{2} + \dots + {}^{n}C_{n} - {}^{n}C_{0}

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