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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 : 39 of 120

Marks: +1, -0

What is 1 - T1 + 2T2 - 3T3 + ... + (-1)nnTn equal to ?

Solution:

Concept:

The expression is derived from the binomial expansion

(1+x)^n = 1 + T_1 x + T_2 x^2 + \dots + T_n x^n

, where

T_k = \binom{n}{k}

Explanation:

We need to evaluate

S = 1 - T_1 + 2T_2 - 3T_3 + \dots + (-1)^n n T_n

Take a specific value, say

n=3

. Then the binomial expansion gives:

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

Comparing with

1 + T_1x + T_2x^2 + T_3x^3

, we get

T_1=3

T_2=3

T_3=1

Now substitute into the required sum:

S = 1 - 3 + 2\times 3 - 3\times 1 = 1 - 3 + 6 - 3 = 1

Thus for

n=3

, the sum equals

1

The same result holds for all positive integers

n \ge 2

. Hence the sum is

1

Answer:

Option D:

1

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