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Higher Math
Example 8. Expand ( x + y)5 and from that expansion find (3 + 2 x) 5 . 5 5 5 5 Solution : ( x + y ) 5 = x 5 + ⎛⎜ ⎞⎟ x 4 y + ⎛⎜ ⎞⎟ x 3 y 2 + ⎛⎜ ⎞⎟ x 2 y 3 + ⎛⎜ ⎞⎟ xy 4 + y 5 1 2 3 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ 4⎠ 5⋅ 4 3 2 5⋅ 4⋅3 2 3 5⋅ 4 ⋅3⋅ 2 4 x y + x y + xy + y 5 = x 5 + 5x 4 y + 1⋅ 2 1⋅ 2 ⋅ 3 1⋅ 2 ⋅ 3 ⋅ 4 = x 5 + 5 x 4 y + 10 x 3 y 2 + 10 x 2 y 3 + 5 xy 4 + y 5 .
∴ The required expasion: ( x + y)5 = x5 + 5x4 y + 10x3 y 2 + 10x2 y3 + 5xy 4 + y5. Now put x = 3 and y = 2x (2 + 2 x) 5 = 35 + 5 ⋅ 3 4 ⋅ 2 x + 10 ⋅ 33 (2 x) 2 + 10 ⋅ 3 2 (2 x) 3 + 5 ⋅ 3(2 x) 4 + (2 x) 5 . = 243 + 810 x + 1080 x 2 + 720 x 3 + 240 x 4 + 32 x 5 There fore, (3 + 2x)5 = 243 + 810x + 1080x2 + 720x3 + 240x4 + 32x5 6
Example 9. Expand ⎛⎜ x + 12 ⎞⎟ upto fourth term in decending power of x and identify ⎝
x ⎠
the term which is free from x. Solution : We get by binomial theorem, 6 2 3 1 ⎞ ⎛ ⎛ 6 ⎞ x 5 ⎛ 1 ⎞ + ⎛ 6 ⎞ x 4 ⎛ 1 ⎞ + ⎛ 6 ⎞ x 3 ⎛ 1 ⎞ + ................ 6 ( ) x + = x + ⎜ ⎟ ⎜1 ⎟ ⎜ 2 ⎟ ⎜ 2 ⎟ ⎜ 2 ⎟ ⎜ 3 ⎟ ⎜ 2 ⎟ x2 ⎠ ⎝ ⎠ ⎝x ⎠ ⎝ ⎠ ⎝x ⎠ ⎝ ⎠ ⎝x ⎠ ⎝ 6⋅5 4 1 6⋅5⋅4 3 1 x ⋅ 4+ x + ....................... = x6 + 6x3 + 1⋅ 2 1⋅ 2 ⋅ 3 x 6 x 1 = x 6 + 6 x 3 + 15 + 20 3 + ....................... x 20 The required expansion x6 + 6x3 + 15 + 3 + ....................... x And term free from x is 15 7
x⎞ ⎛ Example 10. Expand ⎜ 2 − ⎟ upto first four term in asending power of x. Find also 2⎠ ⎝ 7 (1 ⋅ 995) upto four decimal places. Solution : 7 2 3 x⎞ ⎛ 7⎞ 6 ⎛ x ⎞ ⎛7⎞ 5 ⎛ x ⎞ 7⎞ 4 ⎛ x ⎞ ⎛ ⎛ 7 ⎜ 2 − ⎟ = 2 + ⎜1 ⎟2 ⎜ − ⎟ + ⎜ 2 ⎟2 ⎜ − ⎟ + ⎜ 3 ⎟2 ⎜ − ⎟ + ................ 2⎠ ⎝ ⎠ ⎝ 2⎠ ⎝ ⎠ ⎝ 2⎠ ⎝ ⎠ ⎝ 2⎠ ⎝ 3 x2 7 ⋅ 6 ⋅ 5 ⎛ x ⎞ ⎛ x⎞ 7⋅6 ⋅ 32 ⋅ + ⋅16⎜ − ⎟ + ............. = 128 + 7 ⋅ 64⎜ − ⎟ + 4 1⋅ 2 ⋅ 3 ⎝ 2 ⎠ ⎝ 2 ⎠ 1⋅ 2 7
x⎞ ⎛ ∴ ⎜ 2 − ⎟ = 128 − 224x + 168x2 − 70x3 + ....................... 2⎠ ⎝