The coefficient of x¹2} in the expansion of (x²+2x+2)^8 is

Correct Answer is : 2576\; 25762576

Correct Answer is : 2576\; 25762576

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TG EAMCET 3 May 2025 Shift 2 Paper

Question : 20 of 160

Marks: +1, -0

x^{12}

(x^2+2x+2)^8

Solution:

We need to find the coefficient of

x^{12}

in the expansion of

(x^2+2x+2)^8

Step 1: General Term Formula

Each term in the expansion is made by choosing

P_1

times

x^2, P_2

times

2x

, and

P_3

times 2 from the 8 factors, so

P_1+P_2+P_3=8

. The general term is:

\; \frac{8!}{P_1!P_2!P_3!} (x^2)^{P_1} (2x)^{P_2} (2)^{P_3}

Step 2: Power of

x

The power of

x

in each term is

2P_1

(from

x^2

) plus

P_2

(from

2x

), for a total of

2P_1+P_2

Step 3: Find Values for

x^{12}

We want

2P_1+P_2=12

and

P_1+P_2+P_3=8

Possible solutions are:

Step 4: Write Each Term's Coefficient

For each solution, we substitute into the formula:

When

P_1=6, P_2=0, P_3=2

Coefficient:

\; \frac{8!}{6!0!2!} \cdot 2^0 \cdot 2^0 \cdot 2^2

When

P_1=5, P_2=2, P_3=1

Coefficient:

\; \frac{8!}{5!2!1!} \cdot 2^2 \cdot 2^1

When

P_1=4, P_2=4, P_3=0

Coefficient:

\; \frac{8!}{4!4!0!} \cdot 2^4

Step 5: Calculate Coefficients

Total Coefficient

\; = \; \frac{8!}{6!2!} \cdot 2^2 + \; \frac{8!}{5!2!1!} \cdot 2^2 \cdot 2^1 + \; \frac{8!}{4!4!} \cdot 2^4

\; = 28 \times 4 + 168 \times 4 \times 2 + 70 \times 16

\; = 112 + 1344 + 1120

\; = 2576

The coefficient of

x^{12}

is 2576 .

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