Question

Using binomial theorem evaluate each of the following:

$\left(i\right)\left(96{)}^3\right(ii\left)\right(102{)}^5\left(iii\right)\left(101{)}^4\right(iv\left)\right(98{)}^5$

Answer 1

$(96{)}^3$

We have,

$(96{)}^3=(100-4{)}^3$

${=}^3{C}_0\times {100}^3{+}^3{C}_1\times {100}^2\times (-4){+}^3{C}_2\times 100\times (-4{)}^2{+}^3{C}_3\times (-4{)}^3$

$={100}^3-3\times {100}^2\times 4+3\times 100\times 4^2-4^3$

=1000000 -120000+4800-64

=1004800-120064

=884736

$\therefore (96{)}^3=884736$

$\left(ii\right)(102{)}^5$

We have,

$(102{)}^5=(100+2{)}^5$

${=}^5{C}_0\times {100}^5{+}^5{C}_1\times {100}^4\times 2{+}^5{C}_2\times {100}^3\times 2^2{+}^5{C}_3\times {100}^2\times 2^3{+}^5{C}_4\times 100\times 2^4{+}^5{C}_5\times 2^5$

$={100}^5+5\times {100}^4\times 2+10\times {100}^3\times 2^2+10\times {100}^2\times 2^3+5\times 100\times 2^4+2^5$

=10000000000+1000000000+40000000+800000+8000+32

=11040808032

$\therefore (102{)}^5=11040808032$

$\left(iii\right)(101{)}^4$

We have,

$(101{)}^4=(100+1{)}^4$

${=}^4{C}_0\times {100}^4{+}^4{C}_1\times {100}^3{\times }^4{C}_2\times {100}^2{+}^4{C}_3\times 100{+}^4{C}_4$

$={100}^4+4\times {100}^3+6\times {100}^2+4\times 100+1$

=100000000+4000000+60000+400+1

=104060401

$\therefore (101{)}^4=104060401$

$\left(iv\right)(98{)}^5$

We have,

${=}^5{C}_0\times {100}^5{+}^5{C}_1\times {100}^4\times (-2){+}^5{C}_2\times {100}^3+(-2{)}^2{+}^5{C}_3\times {100}^2\times (-2{)}^3{+}^5{C}_4\times 100\times (-2{)}^4{}^{}{+}^5{C}_5\times (-2{)}^5$

${=}^5{C}_0\times {100}^5{-}^5{C}_1\times {100}^4\times 2{+}^5{C}_2\times {100}^3\times 4{-}^5{C}_3\times {100}^2\times 8{+}^5{C}_4\times 100\times 16{-}^5{C}_5\times 32$

$={100}^5-10\times {100}^4+40\times {100}^3-80\times {100}^2+80\times 100-32$

=10000000000-1000000000+40000000-800000+8000-32

=10040008000-1000800032

=9039207968

$\therefore (98{)}^5=9039207968$