The coefficient of x and x\(^2\) in (1 + x)\(^p\) (1 – x)\(^q\) are 4 and – 5, then 2p + 3q is
Correct Answer: 63
Solution and Explanation
The correct answer is 63
\((1+x)^p(1-x)^q\)
\((1+px+\frac{p(p-1)}{2!}x^2+......)\)
\((1-qx+\frac{q(q-1)}{2!}x^2-......)\)
\(p-q=4\)
\(\frac{p(p-1)}{2}+\frac{q(q-1)}{2}-pq=-5\)
p2+q2 - p - 2 - 2pq=-10
(q+4)2 + q2 - (q-4) - q - 2(4+q)q = -10
q2 + 8q + 16 - q2 - q - 4 - q - 8q - 2q2 = -10
-2q=-22
so , q=11 and p=15
\(\therefore \) 2p + 3q = 2(15) + 3(11)
= 30 + 33
= 63
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Concepts Used:
Binomial Expansion Formula
The binomial expansion formula involves binomial coefficients which are of the form
(n/k)(or) nCk and it is calculated using the formula, nCk =n! / [(n - k)! k!]. The binomial expansion formula is also known as the binomial theorem. Here are the binomial expansion formulas.
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. This formula says:
We have (x + y)n = nC0 xn + nC1 xn-1 . y + nC2 xn-2 . y2 + … + nCn yn
General Term = Tr+1 = nCr xn-r . yr
- General Term in (1 + x)n is nCr xr
- In the binomial expansion of (x + y)n , the rth term from end is (n – r + 2)th .