NDA 2020 Maths · Mathematics · Binomial Theorem · medium

Binomn0+binomn1+…+binomnn=2^n. 2. binomn0+binomn2+binomn4+…=2^n-1 for every positive integer n.

Single correct option

Consider the following statements:
1.

\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}=2^n

.
2.

\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots=2^{n-1}

for every positive integer

n

.
Which of the statements is/are correct?

Show answer and solution

Answer: (C) Both 1 and 2

Putting

x=1

(1+x)^n

gives statement 1. Putting

x=-1

gives alternating sum zero, so even-indexed and odd-indexed sums are equal, each being

2^{n}/2=2^{n-1}

for

n\ge1

. Both are correct; for

n=10

the even sum is 512.

Original question written in the NDA 2020 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake