according to ques,
Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer
To show that $$ \[9n+1\text{ }\text{ }8n\text{ }\text{ }9\] is divisible by\[64\] , it must be shown that\[9n+1\text{ }\text{ }8n\text{ }\text{ }9\text{ }=\text{ }64\text{ }k\] , where k is some...
Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate
Utilizing binomial hypothesis the articulations, $$ \[\left( x\text{ }+\text{ }1 \right)6\] and \[\left( x\text{ }\text{ }1 \right)6\] can be communicated as \[\left( x\text{ }+\text{ }1...
Find (a + b)4 – (a – b)4. Hence, evaluate
Utilizing binomial hypothesis the articulation \[\left( a\text{ }+\text{ }b \right)4\] and\[\left( a\text{ }\text{ }b \right)4\] , can be extended \[\left( a\text{ }+\text{ }b \right)4\text{...
Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000
By parting the given $$ \[1.1\] and afterward applying binomial hypothesis, the initial not many terms of \[\left( 1.1 \right)10000\] can be acquired as \[\left( 1.1 \right)10000\text{ }=\text{...
solve (99)5
Given \[\left( 99 \right)5\] \[99\] can be composed as the aggregate or distinction of two numbers then binomial hypothesis can be applied. The given inquiry can be composed as \[99\text{ }=\text{...
solve the following
(101)4 Given (101)4 \[101\] can be communicated as the total or distinction of two numbers and afterward binomial hypothesis can be applied. The given inquiry can be composed as (101)4 = (100 + 1)4...
7. Solve (102)5
Given $$ \[\left( 102 \right)5\] \[102\] can be communicated as the total or contrast of two numbers. now we use binomial theorem, The given inquiry can be composed as \[102\text{ }=\text{...
(96)3
Given \[\left( 96 \right)3\] \[96\] can be communicated as the aggregate or contrast of two numbers and afterward binomial hypothesis can be applied. The given inquiry can be composed as \[96\text{...
Expand each of the expressions in Exercises 1 to 5.
From binomial hypothesis, given condition can be extended as
Expand each of the expressions in Exercises 1 to 5.
From binomial hypothesis, given condition can be extended as
Expand each of the expressions in Exercises 1 to 5.
\[\mathbf{3}.\text{ }{{\left( \mathbf{2x}\text{ }\text{ }\mathbf{3} \right)}^{\mathbf{6}}}\] From binomial hypothesis, given condition can be extended as
Expand each of the expressions in Exercises 1 to 5.
From binomial hypothesis extension we can compose as
Expand each of the expressions in Exercises 1 to 5.
(1 – 2x)5 Solution: From binomial theorem expansion we can write as (1 – 2x)5 = 5Co (1)5 – 5C1 (1)4 (2x) + 5C2 (1)3 (2x)2 – 5C3 (1)2 (2x)3 + 5C4 (1)1 (2x)4 – 5C5 (2x)5 = 1 – 5 (2x) + 10 (4x)2 – 10...