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Exercise 7.1 · Q1

Q.Expand the expression (1−2x)5(1 - 2x)^5.

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The key idea is to apply the Binomial Theorem to (1−2x)5(1 - 2x)^5, treating 11 as aa and −2x-2x as bb. The expanded form is 1−10x+40x2−80x3+80x4−32x51 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5.

The Binomial Theorem gives us a clean way to expand any expression of the form (a+b)n(a + b)^n without having to multiply it out step by step. For a positive integer nn, the theorem states:

(a+b)n=∑k=0n(nk)an−kbk(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

Here, (nk)\binom{n}{k} is the binomial coefficient, often read as "n choose k", and it equals n!k!(n−k)!\frac{n!}{k!(n-k)!}. The pattern is simple: the first term has ana^n (when k=0k=0), the last term has bnb^n (when k=nk=n), and in between, the powers of aa decrease while the powers of bb increase, each multiplied by the appropriate coefficient.

In our problem, we have (1−2x)5(1 - 2x)^5. Compare this to (a+b)n(a + b)^n:

  • a=1a = 1
  • b=−2xb = -2x (notice the negative sign is part of bb)
  • n=5n = 5

So we will expand using the formula with a=1a=1, b=−2xb=-2x, and n=5n=5.

  1. Write the general term. The kk-th term in the expansion (starting from k=0k=0) is:

Tk+1=(5k)(1)5−k(−2x)kT_{k+1} = \binom{5}{k} (1)^{5-k} (-2x)^k

Since 11 raised to any power is just 11, this simplifies to:

Tk+1=(5k)(−2x)kT_{k+1} = \binom{5}{k} (-2x)^k

  1. Compute each term for k=0,1,2,3,4,5k = 0, 1, 2, 3, 4, 5.

    We'll need the binomial coefficients (5k)\binom{5}{k}. These are:

    • (50)=1\binom{5}{0} = 1
    • (51)=5\binom{5}{1} = 5
    • (52)=10\binom{5}{2} = 10
    • (53)=10\binom{5}{3} = 10
    • (54)=5\binom{5}{4} = 5
    • (55)=1\binom{5}{5} = 1

    Now, for each kk, compute (−2x)k(-2x)^k and multiply by the coefficient.

    • k=0k=0: (50)(−2x)0=1⋅1=1\binom{5}{0} (-2x)^0 = 1 \cdot 1 = 1
    • k=1k=1: (51)(−2x)1=5⋅(−2x)=−10x\binom{5}{1} (-2x)^1 = 5 \cdot (-2x) = -10x
    • k=2k=2: (52)(−2x)2=10⋅(4x2)=40x2\binom{5}{2} (-2x)^2 = 10 \cdot (4x^2) = 40x^2 (Remember: (−2)2=4(-2)^2 = 4)
    • k=3k=3: (53)(−2x)3=10⋅(−8x3)=−80x3\binom{5}{3} (-2x)^3 = 10 \cdot (-8x^3) = -80x^3 ((−2)3=−8(-2)^3 = -8)
    • k=4k=4: (54)(−2x)4=5⋅(16x4)=80x4\binom{5}{4} (-2x)^4 = 5 \cdot (16x^4) = 80x^4 ((−2)4=16(-2)^4 = 16)
    • k=5k=5: (55)(−2x)5=1⋅(−32x5)=−32x5\binom{5}{5} (-2x)^5 = 1 \cdot (-32x^5) = -32x^5 ((−2)5=−32(-2)^5 = -32)
  2. Add all the terms together.

    The expansion is the sum of these six terms:

1+(−10x)+40x2+(−80x3)+80x4+(−32x5)1 + (-10x) + 40x^2 + (-80x^3) + 80x^4 + (-32x^5)

So the final expanded form is:

1−10x+40x2−80x3+80x4−32x51 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5

Watch out

A common mistake is forgetting that b=−2xb = -2x, not 2x2x. If you treat bb as 2x2x, the signs of the odd-powered terms (like xx, x3x^3, x5x^5) will be wrong. Always include the negative sign inside the parentheses when raising to a power.

Tip

Notice the symmetry in the coefficients: 1,5,10,10,5,11, 5, 10, 10, 5, 1 from Pascal's triangle. The signs alternate because (−2x)k(-2x)^k gives a positive sign when kk is even and a negative sign when kk is odd. This pattern helps you check your work quickly.

✓Final answer

The expanded form is 1−10x+40x2−80x3+80x4−32x5\boxed{1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5}.

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