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Binomial Theorem question

2023 · 15 Apr · Shift 1 · Q31

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Mathematics
Binomial Theorem
2023 · 15 Apr · Shift 1 · Q31

Binomial Theorem question

2023 · 15 Apr · Shift 1 · Q31

JEE MainMathematicsBinomial TheoremMCQ+4 / −1

Let

\left(a+b x+c x^{2}\right)^{10}=\sum\limits_{i=0}^{20} p_{i} x^{i}, a, b, c \in \mathbb{N}

. If

p_{1}=20

and

p_{2}=210

, then

2(a+b+c)

is equal to :

A
15
B
8
C
6
D
12

View written solutionFree

Correct answer: D

We have

(a+bx+cx^2)^{10}= \sum_{i=0}^{20} p_i x^i.

We are given: $p_1=20, \qquad p_2=210.$

We need to find: $2(a+b+c).$

Find $p_1$

To get the coefficient of $x^1$ in $(a+bx+cx^2)^{10}$ , we must choose $bx$ from exactly one factor and $a$ from the remaining $9$ factors.

Hence,

p_1=\binom{10}{1} a^9 b=10a^9b.

Given $p_1=20$ , we get

Since $a,b\in \mathbb N$ , the only possibility is

Find $p_2$

The coefficient of $x^2$ comes from two cases:

choosing $bx$ from two factors and $a$ from the remaining $8$ factors,
choosing $cx^2$ from one factor and $a$ from the remaining $9$ factors.

So,

p_2=\binom{10}{2}a^8b^2+\binom{10}{1}a^9c.

Substitute $a=1, b=2$ :

p_2=\binom{10}{2}(2)^2+10c=45\cdot 4+10c=180+10c.

Given $p_2=210$ ,

Compute $2(a+b+c)$

Now,

Therefore,

Check options

The value is $12,$ which corresponds to Option D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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