If in the expansion of (1+x)p(1–x)q, the coefficients of x and x2 are 1 and –2, respectively, then p2+q2 is equal to : [2025]
(4)
(1+x)p(1–x)q=(C0p+C1x+C2px2+...)(C0q-C1qx+C2px2+...)
Coeff. of x≡–C0pC1q+C1pC0q=1
⇒ p–q=1 ... (i)
Coeff. of x2≡C0pC2q–C1pC1q+C2pC0q=–2
⇒ q(q–1)2–pq+p(p–1)2=–2
⇒ q2–q–2pq+p2–p=–4
⇒ 1–(p+q)=–4 [∵ q2+p2–2pq=(p–q)2]
⇒ p+q=5 ... (ii)
On solving equation (i) and (ii), we get p = 3 and q = 2.
So, .p2+q2=13