Binomial Theorem, Class 11 Mathematics R.D Sharma Question Answer
Page 18.11 Ex 18.1 Q1. Answer : (i) (2x + 3y)5 =C05(2x)5(3y)0+C15(2x)4(3y)1+C25(2x)3(3y)2+C35(2x)2(3y)3+C45(2x)1(3y)4+C55(2x)0(3y)5 =32×5+5×16×4×3y+10×8×3×9y2+10×4×2×27y3+5×2x×81y4+243y5=32×5+240x4y+720x3y2+1080x2y3+810xy4+243y5 (ii) (2x − 3y)4 =C04(2x)4(3y)0-C14(2x)3(3y)1+C24(2x)2(3y)2-C34(2x)1(3y)3+C44(2x)0(3y)4=16×4-4×8×3×3y+6×4×2×9y2-4×2x×27y3+81y4=16×4-96x3y+216x2y2-216xy3+81y4 (iii) x-1×6=C06 x61x0-C16 x51x1+C26 x41x2-C36 x31x3+C46 x21x4-6C5 x11x5+C66 x01x6=x6-6 x5×1x+15 x4×1×2-20×3×1×3+15×2×1×4-6 x×1×5+1×6=x6-6×4+15×2-20+15×2-6×4+1×6 (iv) (1 − 3x)7 =C07(3x)0-C17(3x)1+C27(3x)2-C37(3x)3+C47(3x)4-C57(3x)5+C67(3x)6-C77(3x)7=1-7×3x+21×9×2-35×27×3+35×81×4-21×243×5+7×729×6-2187×7=1-21x+189×2-945×3+2835×4-5103×5+5103×6-2187×7 (v) (ax-bx)6=C06(ax)6(bx)0-C16(ax)5(bx)1+C26(ax)4(bx)2-C36(ax)3(bx)3+C46(ax)2(bx)4-C56(ax)1(bx)5+C66(ax)0(bx)6 =a6x6-6a5x5×bx+15a4x4×b2x2-20a3b3×b3x3+15a2x2×b4x4-6ax×b5x5+b6x6=a6x6-6a5x4b+15a4x2b2-20a3b3+15a2b4x2-6ab5x4+b6x6 (vi) xa-ax6=C06xa6ax0-C16xa5ax1+C26xa4ax2-C36xa3ax3+C46xa2ax4-C56xa1ax5+C66xa0ax6=x3a3-6x2a2+15xa-20+15ax-6a2x2+a3x3 (vii) x3-a36=C06(x3)6(a3)0-C16(x3)5(a3)1+C26(x3)4(a3)2-C36(x3)3(a3)3+C46(x3)2(a3)4-C56(x3)1(a3)5+C66(x3)0(a3)6=x2-6×5/3a1/3+15×4/3a2/3-20xa+15×2/3a4/3-6×1/3a5/3+a2 (viii) (1+2x-3×2)5Consider 1-2x and 3×2 as two separate entities and apply the binomial […]