Question

For a sequence of real numbers \(x_1, x_2, ..., x_n,\) if \(x_1 - x_2 + x_3 - ... + (-1)^{n + 1}x_n =n^2 + 2n\) for all natural numbers n, then the sum \(x_{49} + x_{50}\) equals

• A. 2
• B. -2
• C. 200
• D. -200

Answer 1

The Correct Option is B

From the given series we have:\(x_1=1+2=3\)
Now \(x_1-x_2=8\)
\(⇒ x=-5\)

We have : \(x_1-x_2+x_3=15\)
\(⇒ x_3=7\)
So,we have: \(x_n=(-1)^{n+1}(2n+1)\)
\(⇒ x_{49}=99 \)
\(⇒ x_{50}=-101\)
\(∴ x_{49}+x_{50}=-2\)

Answer 2

Given that a generic formula in n for the sum of first n terms.
\(S_n = x_1 - x_2 + x_3 - x_4 + ….. + (-1)^{n+1} x_n = n^2 + 2n = n(n + 2)\)
Even terms have a negative sign, while the odd terms have a positive sign.
\(S_{50} = 50 (50 + 2) = 50 (52) = 2600\)
\(S_{49} = 49 (49 + 2) = 49 (51) = 2499\)
\(S_{48} = 48 (48 + 2) = 48 (50) = 2400\)
\(S_{50} = S_{49} - x_{50}\)
\(x_{50} = S_{49} - S_{50} = 2499 - 2600 = -101\)
\(S_{49} = S_{48} + x_{49}\)
\(x_{49}= S_{49}- S_{48} = 2499 - 2400 = 99\)
Now, 
\(x_{49} + x_{50} = 99 - 101 = -2\)

So, the correct option is (B): \(-2\)