We can express 7103 as 7 -50-151-7-50 k -1-A- TrueB- False

The correct option is A True 7^103 = 7 × (49)^51 = 7 × (50 1)^51 7 × (50 1)^51 = 7 [^51C_0(50)^51 ^51C_1(50)^50 + ^51C_2(50)^49............ ^51C_50(50)^ ...

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Byju's Answer

Standard VII

Mathematics

Exponent

We can expres...

Question

We can express $7^{103}$ as $7 \times (50 - 1)^{51} = 7 [50 k - 1]$

True

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False

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Solution

The correct option is A
True

$7^{103}$ = 7 $\times (49)^{51}$

= 7 $\times (50 - 1)^{51}$

7 $\times (50 - 1)^{51}$ = $7 [^{51} C_{0} (50)^{51} -^{51} C_{1} (50)^{50} +^{51} C_{2} (50)^{49} . . . . . . . . . . . .^{51} C_{50} (50)^{1} -^{51} C_{51}]$

7 $\times (50 - 1)^{51}$ = $7 [^{51} C_{0} (50)^{51} -^{51} C_{1} (50)^{50} +^{51} C_{2} (50)^{49} . . . . . . . . . . . .^{51} C_{50} (50)^{1} - 1]$

$^{51} C_{0} (50)^{51} -^{51} C_{1} (50)^{50} +^{51} C_{2} (50)^{49} . . . . . . . . . . . .^{51} C_{50} (50)^{1}$ is a multiple of 50.We will replace it with 50k

7 $\times (50 - 1)^{51} = 7 [50 k - 1]$

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We can express 7103 as 7×(50−1)51=7[50k−1]

We can express $7^{103}$ as $7 \times (50 - 1)^{51} = 7 [50 k - 1]$