Let the coefficients of three consecutive terms in the binomial expansion of (1+2x)^n be in the ratio 2 : 5 : 8. Then the coefficient of the term, which is in the middle of those three terms, is \underline\hspace2cm .

Binomial Theorem PYQ — JEE Main 2023 · 29 Jan · Shift 1 · Q43 with Solution

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Correct answer: 1120

Let the three consecutive terms in the expansion of $(1+2x)^n$ be the $(r+1)$ -th, $(r+2)$ -th and $(r+3)$ -th terms.

1. General coefficient in $(1+2x)^n$

The general term is $T_{k+1} = \binom{n}{k}(2x)^k$ So the coefficient of the $(k+1)$ -th term is $\binom{n}{k}2^k$

Hence the three consecutive coefficients are: $\binom{n}{r}2^r, \quad \binom{n}{r+1}2^{r+1}, \quad \binom{n}{r+2}2^{r+2}$

They are in the ratio $2:5:8$ .

2. Use ratios of consecutive coefficients

Let $A=\binom{n}{r}2^r, \quad B=\binom{n}{r+1}2^{r+1}, \quad C=\binom{n}{r+2}2^{r+2}$ Given $A:B:C = 2:5:8$ So, $\frac{B}{A} = \frac{5}{2}, \qquad \frac{C}{B} = \frac{8}{5}$

Now,

= 2\cdot \frac{n-r}{r+1}$$ Therefore, $$2\cdot \frac{n-r}{r+1} = \frac{5}{2}$$ $$\frac{n-r}{r+1} = \frac{5}{4}$$ $$4n-4r = 5r+5$$ $$4n = 9r+5 \quad ...(1)$$ Also, $$\frac{C}{B} = \frac{\binom{n}{r+2}2^{r+2}}{\binom{n}{r+1}2^{r+1}} = 2\cdot \frac{n-r-1}{r+2}$$ So, $$2\cdot \frac{n-r-1}{r+2} = \frac{8}{5}$$ $$\frac{n-r-1}{r+2} = \frac{4}{5}$$ $$5n-5r-5 = 4r+8$$ $$5n = 9r+13 \quad ...(2)$$ ## 3. Solve for $n$ and $r$ From (1): $$4n = 9r+5$$ From (2): $$5n = 9r+13$$ Subtract appropriately: From (1), $$9r = 4n-5$$ From (2), $$9r = 5n-13$$ So, $$4n-5 = 5n-13$$ $$n=8$$ Then from (1): $$4(8)=9r+5$$ $$32=9r+5$$ $$9r=27$$ $$r=3$$ ## 4. Find the middle coefficient The middle coefficient is $$B=\binom{8}{4}2^4$$ Since $$\binom{8}{4}=70, \qquad 2^4=16$$ we get $$B=70\cdot 16=1120$$ ## 5. Final answer Therefore, the coefficient of the middle term is $$\boxed{1120}$$

More from Binomial Theorem

1. General coefficient in

(1+2x)^n

The general term is

T_{k+1} = \binom{n}{k}(2x)^k

So the coefficient of the

(k+1)

-th term is

\binom{n}{k}2^k

Hence the three consecutive coefficients are:

\binom{n}{r}2^r, \quad \binom{n}{r+1}2^{r+1}, \quad \binom{n}{r+2}2^{r+2}

They are in the ratio

2:5:8

2. Use ratios of consecutive coefficients

Let

A=\binom{n}{r}2^r, \quad B=\binom{n}{r+1}2^{r+1}, \quad C=\binom{n}{r+2}2^{r+2}

Given

A:B:C = 2:5:8

So,

\frac{B}{A} = \frac{5}{2}, \qquad \frac{C}{B} = \frac{8}{5}

Now,

= 2\cdot \frac{n-r}{r+1}$$ Therefore, $$2\cdot \frac{n-r}{r+1} = \frac{5}{2}$$ $$\frac{n-r}{r+1} = \frac{5}{4}$$ $$4n-4r = 5r+5$$ $$4n = 9r+5 \quad ...(1)$$ Also, $$\frac{C}{B} = \frac{\binom{n}{r+2}2^{r+2}}{\binom{n}{r+1}2^{r+1}} = 2\cdot \frac{n-r-1}{r+2}$$ So, $$2\cdot \frac{n-r-1}{r+2} = \frac{8}{5}$$ $$\frac{n-r-1}{r+2} = \frac{4}{5}$$ $$5n-5r-5 = 4r+8$$ $$5n = 9r+13 \quad ...(2)$$ ## 3. Solve for $n$ and $r$ From (1): $$4n = 9r+5$$ From (2): $$5n = 9r+13$$ Subtract appropriately: From (1), $$9r = 4n-5$$ From (2), $$9r = 5n-13$$ So, $$4n-5 = 5n-13$$ $$n=8$$ Then from (1): $$4(8)=9r+5$$ $$32=9r+5$$ $$9r=27$$ $$r=3$$ ## 4. Find the middle coefficient The middle coefficient is $$B=\binom{8}{4}2^4$$ Since $$\binom{8}{4}=70, \qquad 2^4=16$$ we get $$B=70\cdot 16=1120$$ ## 5. Final answer Therefore, the coefficient of the middle term is $$\boxed{1120}$$

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