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Question: Co-efficient of $\alpha^t$ in the expansion of, $(\alpha + p)^{m-1} + (\alpha + p)^{m-2}(\alpha + q)...

Co-efficient of αt\alpha^t in the expansion of, (α+p)m−1+(α+p)m−2(α+q)+(α+p)m−3(α+q)2+......(α+q)m−1(\alpha + p)^{m-1} + (\alpha + p)^{m-2}(\alpha + q) + (\alpha + p)^{m-3}(\alpha + q)^2 + ......(\alpha + q)^{m-1} where α≠−q\alpha \neq -q and p≠qp \neq q is:

A

mCt(pt−qt)p−q\frac{{}^mC_t(p^t - q^t)}{p-q}

B

mCt(pm−t−qm−t)p−q\frac{{}^mC_t(p^{m-t} - q^{m-t})}{p-q}

C

mCt(pt+qt)p−q\frac{{}^mC_t(p^t + q^t)}{p-q}

D

mCt(pm−t+qm−t)p−q\frac{{}^mC_t(p^{m-t} + q^{m-t})}{p-q}

Answer

The coefficient of αt\alpha^t is mCt(pm−t−qm−t)p−q\frac{{}^mC_t(p^{m-t} - q^{m-t})}{p-q}.

Explanation

Solution

The given expression is a sum of mm terms: S=(α+p)m−1+(α+p)m−2(α+q)+(α+p)m−3(α+q)2+......+(α+q)m−1S = (\alpha + p)^{m-1} + (\alpha + p)^{m-2}(\alpha + q) + (\alpha + p)^{m-3}(\alpha + q)^2 + ......+(\alpha + q)^{m-1}

This is a geometric progression (GP) with:

  1. First term (A): A=(α+p)m−1A = (\alpha + p)^{m-1}
  2. Common ratio (R): R=(α+p)m−2(α+q)(α+p)m−1=α+qα+pR = \frac{(\alpha + p)^{m-2}(\alpha + q)}{(\alpha + p)^{m-1}} = \frac{\alpha + q}{\alpha + p}
  3. Number of terms (N): The powers of (α+p)(\alpha+p) range from (m−1)(m-1) down to 00, and powers of (α+q)(\alpha+q) range from 00 up to (m−1)(m-1). This indicates there are mm terms. So, N=mN = m.

Since p≠qp \neq q, it implies that R=α+qα+p≠1R = \frac{\alpha+q}{\alpha+p} \neq 1. Therefore, we can use the formula for the sum of a finite GP: SN=A(1−RN)1−RS_N = \frac{A(1 - R^N)}{1 - R}.

Substitute the values of A, R, and N into the formula: S=(α+p)m−1(1−(α+qα+p)m)1−α+qα+pS = \frac{(\alpha + p)^{m-1} \left(1 - \left(\frac{\alpha + q}{\alpha + p}\right)^m\right)}{1 - \frac{\alpha + q}{\alpha + p}}

Simplify the numerator: Numerator =(α+p)m−1(1−(α+q)m(α+p)m)= (\alpha + p)^{m-1} \left(1 - \frac{(\alpha + q)^m}{(\alpha + p)^m}\right) =(α+p)m−1((α+p)m−(α+q)m(α+p)m)= (\alpha + p)^{m-1} \left(\frac{(\alpha + p)^m - (\alpha + q)^m}{(\alpha + p)^m}\right) =(α+p)m−(α+q)m(α+p)= \frac{(\alpha + p)^m - (\alpha + q)^m}{(\alpha + p)}

Simplify the denominator: Denominator =1−α+qα+p= 1 - \frac{\alpha + q}{\alpha + p} =(α+p)−(α+q)α+p= \frac{(\alpha + p) - (\alpha + q)}{\alpha + p} =p−qα+p= \frac{p - q}{\alpha + p}

Now, divide the simplified numerator by the simplified denominator: S=(α+p)m−(α+q)mα+pp−qα+pS = \frac{\frac{(\alpha + p)^m - (\alpha + q)^m}{\alpha + p}}{\frac{p - q}{\alpha + p}} S=(α+p)m−(α+q)mp−qS = \frac{(\alpha + p)^m - (\alpha + q)^m}{p - q}

To find the coefficient of αt\alpha^t in this expression, we use the binomial theorem for (α+p)m(\alpha + p)^m and (α+q)m(\alpha + q)^m: (α+p)m=∑k=0mmCkαkpm−k(\alpha + p)^m = \sum_{k=0}^m {}^mC_k \alpha^k p^{m-k} (α+q)m=∑k=0mmCkαkqm−k(\alpha + q)^m = \sum_{k=0}^m {}^mC_k \alpha^k q^{m-k}

Substitute these expansions back into the expression for S: S=1p−q(∑k=0mmCkαkpm−k−∑k=0mmCkαkqm−k)S = \frac{1}{p - q} \left( \sum_{k=0}^m {}^mC_k \alpha^k p^{m-k} - \sum_{k=0}^m {}^mC_k \alpha^k q^{m-k} \right) S=1p−q∑k=0mmCkαk(pm−k−qm−k)S = \frac{1}{p - q} \sum_{k=0}^m {}^mC_k \alpha^k (p^{m-k} - q^{m-k})

To find the coefficient of αt\alpha^t, we set k=tk=t in the summation: Coefficient of αt=1p−qmCt(pm−t−qm−t)\alpha^t = \frac{1}{p - q} {}^mC_t (p^{m-t} - q^{m-t})

Comparing this with the given options, it matches option (B).