If the second, third and fourth terms in the expansion of (x+y)^n are 135, 30 and \frac103 , respectively, then 6(n^3+x^2+y) is equal to \underline\hspace2cm .

Binomial Theorem PYQ — JEE Main 2024 · 6 Apr · Shift 1 · Q53 with Solution

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Correct answer: 806

Let the expansion of $(x+y)^n$ be

(x+y)^n = \sum_{r=0}^n \binom{n}{r}x^{n-r}y^r.

So:

2nd term: $T_2 = \binom{n}{1}x^{n-1}y = nx^{n-1}y = 135$
3rd term: $T_3 = \binom{n}{2}x^{n-2}y^2 = 30$
4th term: $T_4 = \binom{n}{3}x^{n-3}y^3 = \dfrac{10}{3}$

We use ratios of consecutive terms.

1. Ratio of third and second terms

\frac{T_3}{T_2} = \frac{\binom{n}{2}x^{n-2}y^2}{nx^{n-1}y} = \frac{n(n-1)}{2n}\cdot \frac{y}{x} = \frac{n-1}{2}\cdot \frac{y}{x}.

But numerically,

\frac{T_3}{T_2} = \frac{30}{135} = \frac{2}{9}.

Hence,

\frac{n-1}{2}\cdot \frac{y}{x} = \frac{2}{9}. \tag{1}

2. Ratio of fourth and third terms

\frac{T_4}{T_3} = \frac{\binom{n}{3}x^{n-3}y^3}{\binom{n}{2}x^{n-2}y^2} = \frac{n(n-1)(n-2)}{6}\cdot \frac{2}{n(n-1)}\cdot \frac{y}{x} = \frac{n-2}{3}\cdot \frac{y}{x}.

But numerically,

\frac{T_4}{T_3} = \frac{\frac{10}{3}}{30} = \frac{1}{9}.

So,

\frac{n-2}{3}\cdot \frac{y}{x} = \frac{1}{9}. \tag{2}

3. Find $n$

From (1),

\frac{y}{x} = \frac{4}{9(n-1)}.

From (2),

\frac{y}{x} = \frac{1}{3(n-2)}.

Equating,

\frac{4}{9(n-1)} = \frac{1}{3(n-2)}

12(n-2)=9(n-1)

12n-24=9n-9

3n=15 \implies n=5.

4. Find $x$ and $y$

Using $n=5$ in (2):

\frac{5-2}{3}\cdot \frac{y}{x} = \frac{1}{9} \implies \frac{y}{x} = \frac{1}{9}.

So,

y=\frac{x}{9}.

Now use the second term:

T_2 = 5x^4y = 135.

Substitute $y=\frac{x}{9}$ :

5x^4\cdot \frac{x}{9} = 135

\frac{5x^5}{9}=135

5x^5 = 1215

x^5=243=3^5

So,

x=3,

and hence,

y=\frac{3}{9}=\frac{1}{3}.

5. Compute the required value

We need

6\left(n^3+x^2+y\right).

Substitute $n=5$ , $x=3$ , $y=\frac13$ :

6\left(5^3+3^2+\frac13\right) =6\left(125+9+\frac13\right) =6\left(134+\frac13\right) =6\cdot \frac{403}{3} =2\cdot 403 =806.

Therefore, the required integer is

\boxed{806}.

More from Binomial Theorem

View written solutionFree

Correct answer: 806

Let the expansion of $(x+y)^n$ be

(x+y)^n = \sum_{r=0}^n \binom{n}{r}x^{n-r}y^r.

So:

2nd term: $T_2 = \binom{n}{1}x^{n-1}y = nx^{n-1}y = 135$
3rd term: $T_3 = \binom{n}{2}x^{n-2}y^2 = 30$
4th term: $T_4 = \binom{n}{3}x^{n-3}y^3 = \dfrac{10}{3}$

We use ratios of consecutive terms.

1. Ratio of third and second terms

\frac{T_3}{T_2} = \frac{\binom{n}{2}x^{n-2}y^2}{nx^{n-1}y} = \frac{n(n-1)}{2n}\cdot \frac{y}{x} = \frac{n-1}{2}\cdot \frac{y}{x}.

But numerically,

\frac{T_3}{T_2} = \frac{30}{135} = \frac{2}{9}.

Hence,

\frac{n-1}{2}\cdot \frac{y}{x} = \frac{2}{9}. \tag{1}

2. Ratio of fourth and third terms

\frac{T_4}{T_3} = \frac{\binom{n}{3}x^{n-3}y^3}{\binom{n}{2}x^{n-2}y^2} = \frac{n(n-1)(n-2)}{6}\cdot \frac{2}{n(n-1)}\cdot \frac{y}{x} = \frac{n-2}{3}\cdot \frac{y}{x}.

But numerically,

\frac{T_4}{T_3} = \frac{\frac{10}{3}}{30} = \frac{1}{9}.

So,

\frac{n-2}{3}\cdot \frac{y}{x} = \frac{1}{9}. \tag{2}

3. Find $n$

From (1),

\frac{y}{x} = \frac{4}{9(n-1)}.

From (2),

\frac{y}{x} = \frac{1}{3(n-2)}.

Equating,

\frac{4}{9(n-1)} = \frac{1}{3(n-2)}

12(n-2)=9(n-1)

12n-24=9n-9

3n=15 \implies n=5.

4. Find $x$ and $y$

Using $n=5$ in (2):

\frac{5-2}{3}\cdot \frac{y}{x} = \frac{1}{9} \implies \frac{y}{x} = \frac{1}{9}.

So,

y=\frac{x}{9}.

Now use the second term:

T_2 = 5x^4y = 135.

Substitute $y=\frac{x}{9}$ :

5x^4\cdot \frac{x}{9} = 135

\frac{5x^5}{9}=135

5x^5 = 1215

x^5=243=3^5

So,

x=3,

and hence,

y=\frac{3}{9}=\frac{1}{3}.

5. Compute the required value

We need

6\left(n^3+x^2+y\right).

Substitute $n=5$ , $x=3$ , $y=\frac13$ :

6\left(5^3+3^2+\frac13\right) =6\left(125+9+\frac13\right) =6\left(134+\frac13\right) =6\cdot \frac{403}{3} =2\cdot 403 =806.

Therefore, the required integer is

\boxed{806}.

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