Question -
Answer -
(i) (2x + 3y) 5
Let us solve the givenexpression:
(2x + 3y) 5 = 5C0 (2x)5 (3y)0 + 5C1 (2x)4 (3y)1 + 5C2 (2x)3 (3y)2 + 5C3 (2x)2 (3y)3 + 5C4 (2x)1 (3y)4 + 5C5 (2x)0 (3y)5
= 32x5 +5 (16x4) (3y) + 10 (8x3) (9y)2 + 10 (4x)2 (27y)3 +5 (2x) (81y4) + 243y5
= 32x5 +240x4y + 720x3y2 + 1080x2y3 +810xy4 + 243y5
(ii) (2x – 3y) 4
Let us solve the givenexpression:
(2x – 3y) 4 = 4C0 (2x)4 (3y)0 – 4C1 (2x)3 (3y)1 + 4C2 (2x)2 (3y)2 – 4C3 (2x)1 (3y)3 + 4C4 (2x)0 (3y)4
= 16x4 –4 (8x3) (3y) + 6 (4x2) (9y2) – 4 (2x) (27y3)+ 81y4
= 16x4 –96x3y + 216x2y2 – 216xy3 +81y4
(iv) (1 – 3x) 7
Let us solve the givenexpression:
(1 – 3x) 7 = 7C0 (3x)0 – 7C1 (3x)1 + 7C2 (3x)2 – 7C3 (3x)3 + 7C4 (3x)4 – 7C5 (3x)5 + 7C6 (3x)6 – 7C7 (3x)7
= 1 – 7 (3x) + 21 (9x)2 –35 (27x3) + 35 (81x4) – 21 (243x5) + 7 (729x6)– 2187(x7)
= 1 – 21x + 189x2 –945x3 + 2835x4 – 5103x5 + 5103x6 –2187x7
(viii) (1 + 2x – 3x2)5
Let us solve the givenexpression:
Let us consider (1 +2x) and 3x2 as two different entities and apply the binomialtheorem.
(1 + 2x – 3x2)5 = 5C0 (1+ 2x)5 (3x2)0 – 5C1 (1+ 2x)4 (3x2)1 + 5C2 (1+ 2x)3 (3x2)2 – 5C3 (1+ 2x)2 (3x2)3 + 5C4 (1+ 2x)1 (3x2)4 – 5C5 (1+ 2x)0 (3x2)5
= (1 + 2x)5 –5(1 + 2x)4 (3x2) + 10 (1 + 2x)3 (9x4)– 10 (1 + 2x)2 (27x6) + 5 (1 + 2x) (81x8)– 243x10
= 5C0 (2x)0 + 5C1 (2x)1 + 5C2 (2x)2 + 5C3 (2x)3 + 5C4 (2x)4 + 5C5 (2x)5 –15x2 [4C0 (2x)0 + 4C1 (2x)1 + 4C2 (2x)2 + 4C3 (2x)3 + 4C4 (2x)4]+ 90x4 [1 + 8x3 + 6x + 12x2] – 270x6(1+ 4x2 + 4x) + 405x8 + 810x9 –243x10
= 1 + 10x + 40x2 +80x3 + 80x4 + 32x5 – 15x2 –120x3 – 3604 – 480x5 – 240x6 +90x4 + 720x7 + 540x5 + 1080x6 –270x6 – 1080x8 – 1080x7 + 405x8 +810x9 – 243x10
= 1 + 10x + 25x2 –40x3 – 190x4 + 92x5 + 570x6 –360x7 – 675x8 + 810x9 – 243x10
(x) (1 – 2x + 3x2)3
Let us solve the givenexpression:
