In every term, the sum of indices of a and b in the expansion of (a+b)n is

Question

In every term, the sum of indices of a and b in the expansion of (a+b)n is (A) n (B) n+1 (C) n+2 (D) n-1

Tardigrade Admin · Accepted Answer

In the expansion of (a+b)n, the sum of the indices of a and b is n+0=n in the first term, (n-1)+1=n in the second term and so on 0+n=n in the last term. Thus, it can be seen that the sum of the indices of a and b is n in every term of the expansion.

Q. In every term, the sum of indices of $a$ and $b$ in the expansion of $(a + b)^{n}$ is

$n$

$n + 1$

$n + 2$

$n - 1$

In the expansion of $(a + b)^{n}$ , the sum of the indices of $a$ and $b$ is $n + 0 = n$ in the first term, $(n - 1) + 1 = n$ in the second term and so on $0 + n = n$ in the last term.
Thus, it can be seen that the sum of the indices of $a$ and $b$ is $n$ in every term of the expansion.

Q. In every term, the sum of indices of a and b in the expansion of (a+b)n is

n

n+1

n+2

n−1

In the expansion of (a+b)n, the sum of the indices of a and b is n+0=n in the first term, (n−1)+1=n in the second term and so on 0+n=n in the last term. Thus, it can be seen that the sum of the indices of a and b is n in every term of the expansion.

Q. In every term, the sum of indices of $a$ and $b$ in the expansion of $(a + b)^{n}$ is

$n$

$n + 1$

$n + 2$

$n - 1$

In the expansion of $(a + b)^{n}$ , the sum of the indices of $a$ and $b$ is $n + 0 = n$ in the first term, $(n - 1) + 1 = n$ in the second term and so on $0 + n = n$ in the last term.
Thus, it can be seen that the sum of the indices of $a$ and $b$ is $n$ in every term of the expansion.