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UPSC NDA Math Model Paper 5

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Question : 110 of 120

Marks: +1, -0

Find the middle term in the expansion of

(x+3)^{6} ?

$625 x^3$
$625 x^5$
$540 x^5$
$540 x^3$

Solution:

Given:

(x+3)^{6}

Here, n = 6

∵ n = 6 and it as even number

As we know that, in the expansion of

(a+b)^{n}

the middle term is

\left(\frac{n}{2}+1\right)\text{th}

term if

n

is even.

So,

\left(\frac{6}{2}+1\right)\text{th}=4

th term is the middle term in the expansion of

(x+3)^{6}

As we know that, the general term is given by:

T_{r+1} = {}^n C_r \cdot a^{n-r} \cdot b^{r}

Here,

n=6, r=3, a=x

and

b=3

T_4 = T_{(3+1)} = {}^6 C_3 \cdot x^3 \cdot (3)^3 = 540 x^3

Hence, option D is the correct answer.

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