Sum

Find the term containing x^{6} in the expansion of (2 − x) (3x + 1)^{9}

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#### Solution

(2 − x) (3x + 1)^{9} = 2(3x + 1)^{9} − x(3x + 1)^{9}

Consider (3x + 1)^{9 }

Here, a = 3x, b = 1, n = 9

We have, t_{r+1} = ^{n}C_{r} a^{n–r} .b^{r}

= ^{9}C_{r} (3x)^{9–r} .(1)^{r}

= ^{9}C_{r} 3^{9–r} .x^{9–r}

To get the coefficient of x^{6} in 2(3x + 1)^{9}, we must have

x^{9–r} = x^{6}

∴ 9 – r = 6

∴ r = 3

Also, to get the coefficient of x^{6} in x(3x + 1)^{9}, we must have

x.x^{9–r} = x^{6}

∴ x^{10–r} = x^{6}

∴ 10 – r = 6

∴ r = 4

∴ The term containing x^{6} in the expansion of

2(3x + 1)^{9} – x(3x + 1)^{9}

=2 ^{9}C_{r} 3^{9–r} – ^{9}C_{r} 3^{9–r}

= 2 ^{9}C_{3} 3^{9–3} – ^{9}C_{4} 3^{9–4}

= 2 × 84 × (3^{6}) – 126 × 3^{5}

= 2 × 3^{5} [3 × 84 – 63]

= 2 × 243[252 – 63]

= 486 × 189

= 91854

Concept: General Term in Expansion of (a + b)n

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