# Binomial Theorem - Class 11 - Mathematics

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## Extra Questions - Binomial Theorem | NCERT | Mathematics | Class 11

The total number of terms in the expansion of $(x+a)^{47} - (x-a)^{47}$ after simplification is

A) 24

B) 47

C) 48

D) 96

**Solution:**The correct option is **A) 24**

We begin by expanding the two expressions using the binomial theorem:

$$ (x+a)^{47} = \sum_{k=0}^{47} {47 \choose k} x^{47-k} a^k $$

$$ (x-a)^{47} = \sum_{k=0}^{47} {47 \choose k} x^{47-k} (-a)^k = \sum_{k=0}^{47} {47 \choose k} x^{47-k} (-1)^k a^k $$

When we subtract the second expansion from the first, terms where $k$ is even will have coefficients adding up (since $(-1)^k = 1$ when $k$ is even), while terms where $k$ is odd will subtract and cancel out completely:

$$ (x+a)^{47} - (x-a)^{47} = \sum_{k=0}^{47} {47 \choose k} x^{47-k} a^k - \sum_{k=0}^{47} {47 \choose k} x^{47-k} (-1)^k a^k $$

$$ = \sum_{\substack{k=0 \ k \text{ even}}}^{47} 2{47 \choose k} x^{47-k} a^k $$

The transformed equation above sums only the terms where $k$ is even. Since $k$ ranges from $0$ to $47$, the even $k$ values are $0, 2, 4, \ldots, 46$. The number of such terms can be computed by noting that these are equally spaced (step of 2) totals from 0 to 46, inclusive. This constitutes:

$$ \frac{46}{2} + 1 = 24 \text{ terms} $$

Hence, **24 terms** remain in the expansion after simplification, making the correct answer **A) 24**.

The value of $5^{\log _{\frac{1}{5}}\left(\frac{1}{2}\right)} + \log _{\sqrt{2}}\left(\frac{4}{\sqrt{3} + \sqrt{7}}\right) + \log _{\frac{1}{2}}\left(\frac{1}{10 + 2 \sqrt{21}}\right)$ is

A) 3

B) 4

C) 5

D) 6

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Number of terms in the expansion of $(a+b+c+d+e)^{5}$ is

A) 25

B) 63

C) $\mathbf{126}$

D) 252