Question 14 of 20advanced🔧 ApplyLong Answer5 marks

Prove that:

∑r=0n3r nCr=4n\sum_{r=0}^{n} 3^r \, nC_r = 4^n
Correct Answer

Proof:

We need to prove that:

∑r=0n3r nCr=4n\sum_{r=0}^{n} 3^r \, nC_r = 4^n

By the Binomial Theorem for any positive integer nn, we have:

(a+b)n=nC0 an+nC1 an−1b+nC2 an−2b2+⋯+nCn bn(a + b)^n = {}^n C_0 \, a^n + {}^n C_1 \, a^{n-1} b + {}^n C_2 \, a^{n-2} b^2 + \cdots + {}^n C_n \, b^n

This can also be written using summation notation as:

(a+b)n=∑r=0nnCr an−r br(a + b)^n = \sum_{r=0}^{n} {}^n C_r \, a^{n-r} \, b^r

Now, let us substitute a=1a = 1 and b=3b = 3 in the above expansion:

(1+3)n=∑r=0nnCr (1)n−r (3)r(1 + 3)^n = \sum_{r=0}^{n} {}^n C_r \, (1)^{n-r} \, (3)^r

Since (1)n−r=1(1)^{n-r} = 1 for all values of rr, we get:

4n=∑r=0nnCr⋅3r4^n = \sum_{r=0}^{n} {}^n C_r \cdot 3^r

Therefore:

∑r=0n3r nCr=4n\sum_{r=0}^{n} 3^r \, nC_r = 4^n

Hence proved.

Exercise: EXERCISE 7.1 | Q: 14 | (Page 8)
For More Understanding

Explanation

This question tests the application of the Binomial Theorem. The key insight is recognizing that the given expression matches the pattern of a binomial expansion. By substituting a=1a = 1 and b=3b = 3 in the standard formula (a+b)n(a + b)^n, we obtain the required result directly. The context provides the general binomial expansion formula and the summation notation which are essential for this proof.

Solution Steps

  1. Step 1: Recall the Binomial Theorem:

    (a+b)n=∑r=0nnCr an−r br(a + b)^n = \sum_{r=0}^{n} nC_r \, a^{n-r} \, b^r

  2. Step 2: Substitute a=1a = 1 and b=3b = 3 in the expansion:

    (1+3)n=∑r=0nnCr (1)n−r (3)r(1 + 3)^n = \sum_{r=0}^{n} nC_r \, (1)^{n-r} \, (3)^r

  3. Step 3: Simplify (1)n−r=1(1)^{n-r} = 1 to get:

    4n=∑r=0nnCr⋅3r4^n = \sum_{r=0}^{n} nC_r \cdot 3^r

  4. Step 4: Conclude that:

    ∑r=0n3r nCr=4n\sum_{r=0}^{n} 3^r \, nC_r = 4^n

    ∴ Hence proved.

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