Q.Expand the expression $(2x - 3)^6$.

Step 1 — The concept: how many terms a binomial expansion has

By the binomial theorem,

The index $r$ runs over $r = 0, 1, 2, \dots, N$ — that is $N + 1$ values, and each gives one distinct term. So:

(The "+1" is simply because the count starts at $r=0$, not $r=1$ — the commonest slip in this type of question.)

Step 2 — Apply it to the given expansion

For $(2x+3)^{3n}$ the exponent is $N = 3n$. Note that the coefficients 2 and 3 are irrelevant — only the power governs the number of terms. Hence

Step 3 — Set it equal to 22 and solve

Step 4 — Verify

With $n = 7$: the exponent is $3n = 21$, and $(2x+3)^{21}$ expands into $21 + 1 = 22$ terms. ✓

(Distractor check: $n=6$ gives $18+1 = 19$ terms; $n=8$ gives $24+1 = 25$; $n=9$ gives $27+1=28$ — none is 22.)

ANSWER: C

The problem asks for the value of

$^{49}C_3 + ^{48}C_3 + ^{47}C_3 + ^{46}C_3 + ^{45}C_3 + ^{45}C_4$.

At first glance, this is a sum of several binomial coefficients. The last term, $^{45}C_4$, is the odd one out — it has a different bottom index. That’s the key: it pairs naturally with $^{45}C_3$ using Pascal’s identity.

Pascal’s identity says:

$\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$.

So $^{45}C_3 + ^{45}C_4 = ^{46}C_4$.

Now the sum becomes:

$^{49}C_3 + ^{48}C_3 + ^{47}C_3 + ^{46}C_3 + ^{46}C_4$.

Again, $^{46}C_3 + ^{46}C_4 = ^{47}C_4$.

So we have: $^{49}C_3 + ^{48}C_3 + ^{47}C_3 + ^{47}C_4$.

Next, $^{47}C_3 + ^{47}C_4 = ^{48}C_4$.

Now: $^{49}C_3 + ^{48}C_3 + ^{48}C_4$.

Then $^{48}C_3 + ^{48}C_4 = ^{49}C_4$.

So: $^{49}C_3 + ^{49}C_4$.

Finally, $^{49}C_3 + ^{49}C_4 = ^{50}C_4$.

Each step is just applying Pascal’s identity to the last two terms, collapsing the sum from right to left like a telescope.

The key idea is that in a binomial expansion $(a+b)^n$, the $r$-th term (starting with $r=0$ for the first term) is $\binom{n}{r} a^{n-r} b^r$. Here the 6th term corresponds to $r=5$. The coefficient of that term is not just the binomial coefficient — it also includes the numerical factors from $a$ and $b$. Since $a = \frac{1}{x}$ and $b = x \sin x$, the powers of $x$ cancel, leaving only powers of $\sin x$. The given coefficient then becomes an equation in $\sin x$.

1. Identify the term number correctly In $(a+b)^{10}$, the first term has $r=0$, second term $r=1$, …, so the 6th term has $r=5$. The general term is

For the 6th term ($r=5$):

2. Substitute $a = \frac{1}{x}$ and $b = x \sin x$

The $x^5$ cancels completely:

3. Compute the binomial coefficient

So the coefficient of the 6th term is $252 \sin^5 x$.

4. Set equal to the given coefficient The problem states this coefficient equals $7\frac{7}{8} = \frac{63}{8}$.

Divide both sides by 63:

5. Solve for $\sin x$ Taking the fifth root:

The principal value of $x$ (presumably in degrees, as the options suggest) for which $\sin x = \frac12$ is $x = 30^\circ$.

In $(x+a)^{53}+(x-a)^{53}$ the terms with odd powers of $a$ cancel and those with even powers double. For $(x+a)^n+(x-a)^n$ with $n$ odd, the number of surviving terms is $\dfrac{n+1}{2}$. With $n=53$: $\dfrac{54}{2}=27$.