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Consider the expansion of (1 + x)n. Let p, q, r and s be the coefficients of first, second, nth and (n + 1)th terms respectively. What is (ps + qr) equal to?

Consider the expansion of (1 + x)n. Let p, q, r and s be the coefficients of first, second, nth and (n + 1)th terms respectively. What is (ps + qr) equal to? Correct Answer 1 + n<sup>2</sup>

Formula used:

This binomial expansion formula gives the expansion of (x + y)n where 'n' is a natural number. The expansion of (x + y)n has (n + 1) terms. 

(x + y)n = nC 0 xn y0 + nC 1 xn - 1 y1 + nC 2 xn-2 y2 + nC 3 xn - 3 y3 + ... + nC n−1 x yn - 1 + nC n x0yn

Calculation:

(1 + x)n = 1 + n x + ...  + n.xn-1+ xn

From above equation we can write 

p = 1, q = n

r = n, s = 1

Now, We have to find the value of (ps + qr)

⇒  1 + n2

∴ (ps + qr) equal to (1 + n2).

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