Question

Using binomial theorem, find the value of:
i) (102)⁴
ii) (1.1)⁵

Answer 1

please let me know if this answer is correct

Answer 2

Answer:

${(1.1)}^{5} = 1.61051 \\ \\ {(102)}^{4} = 108243216$

Step-by-step explanation:

We know that Binomial Theorem is

${(a + b)}^{n} \\\\= ^{n}C_0 \: {a}^{n} + ^{n}C_1 \: {a}^{n - 1} b + ^{n}C_2 \: {a}^{n - 2} {b}^{2} + ... + ^{n}C_n \: {b}^{n } \\ \\ {(102)}^{4} = {(100 + 2)}^{4} \\\\= ^{4}C_0 \: {(100)}^{4} + ^{4}C_1 \: {100}^{3} (2)+ ^{4}C_2 \: {(100)}^{2} {2}^{2} \\ + ^{4}C_3 \: {(100)}^{1} {2}^{3}+ ^{4}C_4 \: {(100)}^{0} {2}^{4} + \\ \\ = 100000000 + 4 \times 1000000 \times 2 + 6 \times 10000 \times 4 +\\ 4 \times 100 \times 8 + 1 \times 16 \\ \\ = 100000000 + 8000000 + 240000 + 3200 + 16 \\ \\ {(102)}^{4} = 108243216$

${(1.1)}^{5} = {(1 + .1)}^{5} \\\\ = ^{5}C_0 \: {(1)}^{5} + ^{5}C_1 \: {1}^{4} (0.1)+ ^{5}C_2 \: {(1)}^{3} {(0.1)}^{2} \\+ ^{5}C_3 \: {(1)}^{2} {(0.1)}^{3}+ ^{5}C_4 \: {(1)}^{1} {(0.1)}^{4} + ^{5}C_5 \: {(1)}^{0} {(0.1)}^{5}\\ \\ = 1+ 5 \times1\times 0.1 + 10 \times 1 \times 0.01+ 10 \times 1 \times 0.001\\ + 5 \times 1 \times 0.0001 + 1 \times 0.00001 \\ \\ = 1 + 0.5 + 0.1 + 0.01 + 0.0005 + 0.00001\\ \\ {(1.1)}^{5} = 1.61051$

Hope it helps you.