SATHEE: Binomial Theorem 2 Question 11

Binomial Theorem 2 Question 11

14. For any positive integers $m, n$ (with $n \geq m$ ),

If $_{m}^{n} =^{n} C_{m}$ . Prove that

$\frac{n}{m} + \begin{matrix} n - 1 \\ m \end{matrix} + \begin{matrix} n - 2 \\ m \end{matrix} + \dots +_{m}^{m} = \begin{aligned} n + 1 \\ m + 1 \end{aligned}$

Prove that

$\begin{aligned} \begin{array}{c} n \\ m \end{array} + 2 \begin{array}{c} n - 1 \\ m \end{array} + 3_{m}^{n - 2} + \dots + (n - m + 1) \\ m = n + 2 \end{aligned}$

(IIT JEE 2000, 6M)

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Solution:

Let $S =_{m}^{n} +_{m}^{n - 1} +_{m}^{n - 2} + \dots +_{m}^{m} = \begin{aligned} n + 1 & m + 1 \end{aligned}$

It is obvious that, $n \geq m$ .

[given]

NOTE This question is based upon additive loop.

Now, $S =_{m}^{m} +_{m}^{m + 1} +_{m}^{m + 2} + \dots +_{m}^{n}$

$\begin{aligned} = \begin{array}{c} m + 1 \\ m + 1 \end{array} + \begin{array}{c} m + 1 \\ m \end{array} + \begin{array}{c} m + 2 \\ m \end{array} + \dots \begin{array}{c} n \\ m \end{array} \\ ∵_{m}^{m} = 1 = \begin{matrix} m + 1 \\ m + 1 \end{matrix} \\ = \frac{m + 2}{m + 1} + \frac{m + 2}{m} + \dots + \begin{array}{c} n \\ m \end{array} \\ =_{m + 1}^{m + 3} + \dots + \frac{n}{m} \\ = \dots \dots \dots \dots \dots \dots . . . . . . . . . . . . . . . . \\ =_{m + 1}^{n} +_{m}^{n} = \begin{array}{c} n + 1 \\ m + 1 \end{array}, which is true. \end{aligned}$

Again, we have to prove that

$\begin{aligned} \begin{array}{c} n \\ m \end{array} + 2 \begin{array}{c} n - 1 \\ m \end{array} + 3 \begin{array}{c} n - 2 \\ m \end{array} + \dots + (n - m + 1) \begin{matrix} m \\ m \end{matrix} = \begin{matrix} n + 2 \\ m + 2 \end{matrix} \\ Let S_{1} =_{m}^{n} + 2 \begin{array}{cc} n - 1 \\ m \end{array} + 3 \begin{array}{c} n - 2 \\ m \end{array} + \dots + (n - m + 1)_{m}^{m} \\ = \frac{n}{m} + \frac{n - 1}{m} + \frac{n - 2}{m} + \dots + \frac{m}{m} \\ + \begin{array}{c} n - 1 \\ m \end{array} + \frac{n - 2}{m} + \dots + \begin{matrix} m \\ m \end{matrix} \\ + \begin{array}{c} n - 2 \\ m \end{array} + \dots +_{m}^{m} n - m + 1 rows \\ \begin{array}{c} \dots \dots \\ + \begin{matrix} m \\ m \end{matrix} \end{array} \end{aligned}$

Now, sum of the first row is $n + 1$

$m + 1$

Sum of the second row is $\begin{matrix} n m + 1 \end{matrix}$ .

Sum of the third row is $n - 1$

$m + 1$ ,

Sum of the last row is $\begin{aligned} m & m \end{aligned} = \begin{aligned} m + 1 & m + 1 \end{aligned}$ .

Thus, $S = \begin{matrix} n + 1 m + 1 \end{matrix} + \begin{matrix} n m + 1 \end{matrix} + \begin{matrix} n - 1 m + 1 \end{matrix}$

$+ \dots + \begin{matrix} m + 1 \\ m + 1 \end{matrix} = \begin{matrix} n + 1 + 1 \\ m + 2 \end{matrix} = \begin{matrix} n + 2 \\ m + 2 \end{matrix}$

[from Eq. (i) replacing $n$ by $n + 1$ and $m$ by $m + 1$ ]

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Binomial Theorem 2 Question 11

14. For any positive integers $m, n$ (with $n \geq m$ ),

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

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Binomial Theorem 2 Question 11

14. For any positive integers m,n (with n≥m ),

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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14. For any positive integers $m, n$ (with $n \geq m$ ),