Q.Compute

By the binomial theorem, the general term of $(a+b)^n$ is $\binom{n}{k}a^{n-k}b^k$, and the exponents $(n-k)$ and $k$ always add to $n$.

The binomial expansion is $(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}$.

In each term, the exponent of $a$ is $n-k$ and the exponent of $b$ is $k$. Their sum is $(n-k)+k=n$, which is constant across every term.

By the zero-exponent rule, any nonzero quantity raised to the power $0$ equals $1$.

For any real number $k\ne 0$, $k^0 = 1$. Here $k = a + b$ (assuming $a + b\ne 0$).

So $(a+b)^0 = 1$.

The Binomial Theorem expands $(a+b)^n$ as $\sum_{r=0}^{n}{}^{n}C_r\,a^{n-r}b^{r}$.

For a positive integer $n$, the Binomial Theorem states:

$(a+b)^n={}^nC_0a^n+{}^nC_1a^{n-1}b+{}^nC_2a^{n-2}b^2+\cdots+{}^nC_{n-1}ab^{n-1}+{}^nC_nb^n=\displaystyle\sum_{r=0}^{n}{}^{n}C_r\,a^{n-r}b^{r}$

Each coefficient ${}^nC_r=\dfrac{n!}{r!(n-r)!}$ is a binomial coefficient, and the powers of $a$ decrease from $n$ to $0$ while the powers of $b$ increase from $0$ to $n$ across the $(n+1)$ terms.

Every term of $(a+b)^n$ is $^{n}C_{r}\,a^{n-r}b^{r}$, and $(n-r)+r=n$ always.

Step 1. By the Binomial theorem, $(a+b)^n = \sum_{r=0}^{n} {}^{n}C_{r}\,a^{n-r}b^{r}$.

Step 2. In the general term $^{n}C_{r}a^{n-r}b^{r}$, the index of $a$ is $n-r$ and the index of $b$ is $r$.

Step 3. Their sum is $(n-r) + r = n$, a constant for every term.

Writing the coefficients $^nC_0, {}^nC_1, \ldots, {}^nC_n$ row by row for $n=0,1,2,\ldots$ produces Pascal's triangle.

Step 1. For each $n$, the coefficients of $(a+b)^n$ are $^{n}C_{0}, {}^{n}C_{1}, \ldots, {}^{n}C_{n}$.

Step 2. Stacking these rows for $n = 0, 1, 2, \ldots$ produces the triangular array known as Pascal's triangle, where each entry is the sum of the two entries above it.

Number of terms $=n+1$; option (c).

By the Binomial Theorem (NCERT Class 11), $(1+x)^n=\sum_{r=0}^{n}\binom{n}{r}x^r$, and $r$ runs from $0$ to $n$ — that is $n+1$ terms.

The sum of all binomial coefficients of $(a+b)^n$ equals $2^n$.

Putting $a=1, b=1$ in the binomial expansion $(a+b)^n = \sum_{r=0}^{n} {}^nC_r\, a^{n-r}b^r$ gives:

$(1+1)^n = {}^nC_0 + {}^nC_1 + {}^nC_2 + \ldots + {}^nC_n$

$2^n = {}^nC_0 + {}^nC_1 + \ldots + {}^nC_n$

The statement is True.

The general term of the binomial expansion of $(a+b)^n$ is $T_{r+1} = {}^nC_r\, a^{n-r}b^r$.

The exponent of $a$ is $(n-r)$ and the exponent of $b$ is $r$; their sum is $(n-r)+r = n$, for every term.

Substituting $x=1$ into $(1+x)^n = {}^{n}C_0 + {}^{n}C_1x + \dots + {}^{n}C_nx^n$ gives $2^n = {}^{n}C_0+{}^{n}C_1+\dots+{}^{n}C_n$.

By the binomial theorem:

$(1+x)^n = {}^{n}C_0 + {}^{n}C_1x + {}^{n}C_2x^2 + \dots + {}^{n}C_nx^n$

Putting $x = 1$:

$(1+1)^n = {}^{n}C_0 + {}^{n}C_1 + {}^{n}C_2 + \dots + {}^{n}C_n$

$2^n = {}^{n}C_0 + {}^{n}C_1 + \dots + {}^{n}C_n$