1. Solve: (√2 + 1)^5 + (√2 − 1)^5
Solution:
From the general binomial theorem expression, we can write:
(x + y)^5 + (x – y)^5 = 2[5C0 x^5 + 5C2 x^3 y^2 + 5C4 xy^4] where x = √2 and y = 1.
Thus we get,
= 2(x^5 + 10 x3y^2 + 5x y^4);
Now if we replace the values of x and y, we get:
[(√2 + 1)^5 + (√2 − 1)^5 = 2[(√2)^5 + 10(√2)^3(1)^2 + 5(√2)(1)^4]
Which upon further solving gives us our required answer = 58√2
2. Find the middle term in the expansion of 2ax - (b^12/x^2)
Solution
Given binomial expression: 2ax - (b^12/x^2)
Using the properties of binomial terms, since the power to which the terms are raised is even, there is 1 middle term, which is the 7th term of the expanded binomial expression.
Furthermore, we can write:
T7 = 12C6 (2ax)^6 (-b/x^2) ^6
=> T7 = 12C6 (2^6a^6x^6) (-b)^6 / x^12
=> T7 = 12C6 (2^6a^6b^6)/x^6
Further putting in the value of 12C6 we arrive at the result: 59136a^6b^6
Thus, the middle term in the expansion of 2ax - (b^12/x^2) is 59136a^6b^6 (Ans)
3. Show that 2^(4n+ 4) - 15n - 16 is divisible by 225. Assume that n is a Natural number.
Solution:
Given expression is: 2^(4n+ 4) - 15n - 16
Expanding this binomial expression using the properties of binomial theorem, we get:
= 2^4(n+ 1) - 15n - 16
= 16 ^(n+1) - 15n -16
= (1 + 15)^(n +1) - 15n -16
Now if we binominally expand this equation, we get:
[(n+1)C0 15^0 +(n+1) 1C1 15^1 + (n+1) 1C2 15^2 + (n+1) 1C3 15^3 + (n+1) 1C4 15^4 + …. + (n+1) 1C(n+1) 15^(n+1) - 15n -16 ]
Further simplifying we get:
15^2 [(n+1) 1C2 15^2 + (n+1) 1C3 15^3 + (n+1) 1C4 15^4 + ….so on]
To prove that 2^(4n+ 4) - 15n - 16 is divisible by 225, we can check if the above equation in the simplified form is divisible by 225.
Thus, 2^(4n+ 4) - 15n - 16 is divisible by 225.