If the coefficients of x and x2 in the expansion of (1 + x)p (1 – x)q, p, q≤15, are – 3 and – 5 respectively, then coefficient of x3 is equal to ______.
Coefficient of x in (1 + x)p (1 – x)q
\(^{−p}C_0\ ^{q}C_1+ ^{p}C_1\ ^qC_0=−3\)
\(⇒ p−q=−3\)
Coefficient of x2 in (1 + x)p (1 – x)q
\(^pC_0\ ^qC_2− ^pC_1\ ^qC_1 + ^pC_2\ ^qC_0=−5\)
\(\frac{q(q−1)}{2}−pq+\frac{p(p−1)}{2}=−5\)
\(\frac{q^2−q}{2}−(q−3)q+\frac{(q−3)(q−4)}{2}=−5\)
⇒ q = 11, p = 8
Coefficient of x3 in (1 + x)8 (1 – x)11 is
\(=^{−11}C_3+ ^8C_1 ^{11}C_2− ^8C_2 ^{11}C_1+ ^8C_3\)
=23
So, the answer is 23.
Consider the following cell: $ \text{Pt}(s) \, \text{H}_2 (1 \, \text{atm}) | \text{H}^+ (1 \, \text{M}) | \text{Cr}_2\text{O}_7^{2-}, \, \text{Cr}^{3+} | \text{H}^+ (1 \, \text{M}) | \text{Pt}(s) $
Given: $ E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33 \, \text{V}, \quad \left[ \text{Cr}^{3+} \right]^2 / \left[ \text{Cr}_2\text{O}_7^{2-} \right] = 10^{-7} $
At equilibrium: $ \left[ \text{Cr}^{3+} \right]^2 / \left[ \text{Cr}_2\text{O}_7^{2-} \right] = 10^{-7} $
Objective: $ \text{Determine the pH at the cathode where } E_{\text{cell}} = 0. $
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is