Compute $(98)^{5}$

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Solution

$98^{5} = (100 - 2)^{5}$

$= 100^{5} -^{5} C_{1} {.100}^{4} .2 +^{5} C_{2} {.100}^{3} 2^{2} -^{5} C_{3} {.100}^{2} {.2}^{3} +^{5} C_{4} .100 {.2}^{4} -^{5} C_{2}$
$= 100^{5} - {10.100}^{4} + {40.100}^{3} - {80.100}^{2} + 80.100 - 32$
$= {90.100}^{4} + {40.100}^{3} - 80.100 + 8000 - 32$

$= 9000000000 + 40000000 - 300000 + 8000 - 32$

$= 9040000000 - 800000 + 7968$

$= 9039200000 + 7968$

$= 9039207968.$

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The value of - 8765+985

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If sum of the number 985 and two other numbers obtaind by arranging the digits of 985 in cyclic order is divided by 111 , 22 and 37 respectively.

Find the quotient in each case.

If sum of the number $985$ and two other numbers obtained by arranging the digits of $985$ in cyclic order is divided by $111, 22$ and $37$ respectively. Find the quotient in each case.

Q. Using binomial theorem evaluate each of the following:
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Compute (98)5

985=(100−2)5=1005−5C1.1004.2+5C2.100322−5C3.1002.23+5C4.100.24−5C2=1005−10.1004+40.1003−80.1002+80.100−32=90.1004+40.1003−80.100+8000−32=9000000000+40000000−300000+8000−32=9040000000−800000+7968=9039200000+7968=9039207968.

Compute $(98)^{5}$