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UPSC NDA II 2023 Mathematics Solved Paper

Question Numbers: 38-40

Consider the following for the next items that follow:

Let (1 + x)n = 1 + T1x + T2x2 + T3x3 + ... + Tnxn.

Question : 40 of 120

Marks: +1, -0

What is T1 + T2 + T3 + ... + Tn equal to ?

Solution:

Concept:

The sum of the coefficients of

x^1

x^n

in the binomial expansion of

(1+x)^n

2^n - 1

Explanation:

The binomial expansion is:

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

Here,

T_1 = \binom{n}{1}

T_2 = \binom{n}{2}

, …,

T_n = \binom{n}{n}

The sum of all coefficients from

x^0

x^n

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

Subtract the constant term

\binom{n}{0} = 1

to get the required sum:

T_1 + T_2 + \cdots + T_n = 2^n - 1

For verification, take

n=3

(1+x)^3 = 1 + 3x + 3x^2 + x^3

, so

T_1=3

T_2=3

T_3=1

, sum =

7 = 2^3 - 1

This holds for any

n

Answer:

2^n - 1

(Option B or C as given in the paper).

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