Prove  2^2n   -  3n - 1  is  devisible  by  9   for  n>=1?
367524
using principle of mathematical induction u can prove this\n check for n=1  2^2 -3 -1=0 which is div by 9\nlet this be true for n=m\n2^2m -3 m -1 =9k  ,where k is any integer\n\nnow for n=m+1 \n2^2(m+1) -3(m+1) -1\n=4.2^2m-3m-4\n=(2^2m-3m-1)+3.2^2m -3\n=9k+3(2^2m-1), now (2^2m -1) is an odd integer divisible by 3 for all m >0\ntherefore the above expression becomes 9k +3.3g ,where g is any integer\n=9(k+g)=9i  where i is an integer \ntherefore the condition for divisibility by 9 is true for n=m+1\nhence by principle of mathematical induction it is true for all n>=1
269006
Ask your friends if they can find the error in the following proof.\n(it's obviously not right, but it looks cool!)\n\na=b\na+1 = b+1\n(a+1)^2 = (b+1)^2\na^2 + 2a +1 = b^2 + 2b +1\ncancel the ones and rearrange\na^2 - b^2 = 2(b-a)\n(a+b)(a-b) = 2(b-a)\ntherefore a+b = -2\nPick numbers for a and b\nand look!! 5 + 7 = -2\nThus shown.\nLOL
747004
When n = 1 \n1! = 1 ≤ 1^1\n\nAssume true for n = k \nie k! ≤ k^k\n\nWhen n = k + 1\n\n(k+1)! = (k + 1) * k!\n≤ (k+1)*k^k (using the assumption k! ≤ k^k)\n< (k + 1) * (k + 1)^k (as k < k + 1)\n= (k + 1)^(k + 1)\n\nSo if k! ≤ k^k then (k+ 1)! < (k + 1)^(k + 1)\n\nWell 1! ≤ 1^1\nSo (1 + 1)! < (1 + 1)^(1 + 1) ie 2! < 2^2\nSince 2! < 2^2 \nthen (2 + 1)! < (2 + 1)^(2 + 1) ie 3! < 3^3\nSimilarly 4! is less than 4^4 and so forth for all integer values of n greater than or equal to 1\n\nie n! ≤ n^n for all natural number values of n .... QED
954617
2n+1 is an odd integer, the next one is 2n+3 \nso:\n(2n+1) + (2n+3) <= 185\n\n/
787052
1, 1, 9, 9, 25\nFor an odd n, the nth term is n^2.\nFor an even n, the nth term is (n-1)^2.
22440
Well, the power is of course 99, but if you are looking for the coefficient, it is -5050.  Since terms that are constant multiples of X^99 must be multiples of all X's except for one, one simply adds the second terms in each binomial {-1,-2,...,-100}.  The sum is -5050
886343
1/9(x-1)^9 evaluated from 0 to 2.\n\n1/9(1)^9 - 1/9(-1)^9 = 1/9 + 1/9 = 2/9\n\nIf I didn't screw up.
186591
solve it by induction...\nlet n=1, then 5^2n-1 +1 =6 so divisible by 6\nlet n=k, then 5^2n-1 +1=5^2k-1 +1 and let it be divisible by 6\nlet n=k+1 then 5^2n-1 +1=5^(2(k+1)-1)) +1\nie 5^(2k+2-2) +1=(5^2k-1)(5^1)\nbut we know that 5^2k-1 is a multiple of 6,so (5^2n-1)*5 is also a multiple od 6. since it satisfies this for n=0, it should satisfy for all n
990505
See my answer in your next question, Rajesh. \nYour answer only works for powers of 2. (2ⁿ)\n\nAnd I agree with the other guy.  (9/2)² does not yield a perfect square.  What you're doing is taking 2^5 = 9^2 - 7^2 and dividing everything by 2^2 and saying you have a "new solution" to 2^3 = c²-b².
745065
(3m/n)(n^2/9) \n=(3mn^2)/(9n)\n=(3mn)/(9)\n=(mn)/3
667985
You can factor this by grouping:\n\nx^3 + 2x^2 - 9x = 18 \nx^3 + 2x^2 - 9x -18 = 0\nx^2(x + 2) + (-9)(x + 2) = 0\n(x^2 -9)(x + 2) = 0\n(x - 3)(x + 3)(x + 2) = 0\n\nx = { 3, -3, -2}\n\nFirst subtract 18 from both sides, then facter each binomial, and finally use the difference of squares.
358011
let x=45.\n\n2x+1=91=7*13\n\n91 is not divisible by 3 or 5, nor is it a prime, nor a square of a prime.\n\nAny product of two distinct primes greater than 5 will be a counterexample to your claim.
945958
(2n-1) * (2n-2) * (2n-3) * (2n-4)... * 3 * 2 * 1\n\nJust like any other factorial.
779778
Think of the 8 as 2^3.\nThen you have 2^100/2^3 which is 2^97 exactly.(because you subtract the powers from each other)\n\nThere is no remainder.\n\n\nz
931054
Group all elements with y together:\n\ny(x^2 + 6x +9) -x^2 + 2x + 3 = 0\n\nMove the rest to the other side:\n\ny(x+3)^2 = x^2 - 2x + 3 (1)\n\nx = -3 is obviously not the solution of this equation therefore we can divide both sides of (1) by (x+3)^2.\n\n(1) is equivalent to:\n\ny = (x^2 - 2x + 3) / (x+3)^2
766061
p^9-q^9/p^6-q^6\n=(p^3-q^3)(p^6+p^3q^3+q^6)/(p^3+q^3)(p^3-q^3)\n=(p^6+p^3q^3+q^6)/(p^3+q^3)
822221
Using mods...\n\nn^3 + n - 4 = 0 mod 7\n\nn^3 + n = 4 mod 7\n\nn(n^2+1) = 4 mod 7\n\nIf n=0 mod 7 then n(n^2+1) = 0 mod 7, doesn't work.\nIf n=1 mod 7 then n(n^2+1) = 1(1^2+1) mod 7 = 2 mod 7, doesn't work.\nIf n=2 mod 7 then n(n^2+1) = 2(4+1) mod 7 = 3 mod 7, doesn't work.\nIf n=3 mod 7 then n(n^2+1) = 3(9+1) mod 7 = 2 mod 7, doesn't work.\nIf n=4 mod 7 then n(n^2+1) = 4(16+1) mod 7 = 5 mod 7, doesn't work.\nIf n=5 mod 7 then n(n^2+1) = 5(25+1) mod 7 = 130 mod 7 = 4 mod 7 - YESSSS!\nIf n=6 mod 7 then n(n^2+1) = 6(36+1) mod 7 = 222 mod 7 = 5 mod 7, doesn't work.\n\nSo, n must be equal to 7m+5 for some integer m.  This gives us:\nn^3 + n - 4 =\n(7m+5)^3 + 7m+5 - 4 =\n7^3m^3 + 3*7^2m^2*5 + 3*7m*5^2 + 5^3 + 7m + 1 = \na whole bunch of multiples of 7 + 5^3 + 1 = \nmultiples of 7, plus 126 =\nmultiples of 7, plus 18*7.