Generalizing $(^4C_0)^2-(^4C_1)^2+(^4C_2)^2-(^4C_3)^2+(^4C_4)^2=,^4C_2$

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I got this question, and I have a bit of an issue with part (b). I managed to prove the attached result; however, it does not really fit the above result (as it said generalise the result and then prove). Can anyone assist?




binomial-coefficients binomial-theorem




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  • You surely want to find $$sum_k=0^n(-1)^knchoose k^2.$$
    – Lord Shark the Unknown
    Aug 16 at 3:54










  • @LordSharktheUnknown I guess so but I can't seem to generate anything similar to that from the given formula that they told me to expand. I tried values for my formula and it works but there is probably something I'm not seeing to get it to look like the above statement
    – user122343
    Aug 16 at 4:39










  • You are generalizing incorrectly. Essentially, consider the expression given in part b, and compute the coefficient of $frac 1x^2n$ on both sides using the binomial theorem. The LHS will give you some expression, and the RHS some expression. These must be the same, and their equality is a generalization of the specific case you solved.
    – Ð°ÑÑ‚он вілла олоф мэллбэрг
    Aug 16 at 6:13










  • Thank you . I was comparing the coefficients for 1x^4 but I understand why you took 2n as it fits 1x^4 when n=2
    – user122343
    Aug 16 at 6:18















up vote
0
down vote

favorite












the question



the answer



I got this question, and I have a bit of an issue with part (b). I managed to prove the attached result; however, it does not really fit the above result (as it said generalise the result and then prove). Can anyone assist?




binomial-coefficients binomial-theorem




share|cite|improve this question






















  • You surely want to find $$sum_k=0^n(-1)^knchoose k^2.$$
    – Lord Shark the Unknown
    Aug 16 at 3:54










  • @LordSharktheUnknown I guess so but I can't seem to generate anything similar to that from the given formula that they told me to expand. I tried values for my formula and it works but there is probably something I'm not seeing to get it to look like the above statement
    – user122343
    Aug 16 at 4:39










  • You are generalizing incorrectly. Essentially, consider the expression given in part b, and compute the coefficient of $frac 1x^2n$ on both sides using the binomial theorem. The LHS will give you some expression, and the RHS some expression. These must be the same, and their equality is a generalization of the specific case you solved.
    – Ð°ÑÑ‚он вілла олоф мэллбэрг
    Aug 16 at 6:13










  • Thank you . I was comparing the coefficients for 1x^4 but I understand why you took 2n as it fits 1x^4 when n=2
    – user122343
    Aug 16 at 6:18













up vote
0
down vote

favorite









up vote
0
down vote

favorite











the question



the answer



I got this question, and I have a bit of an issue with part (b). I managed to prove the attached result; however, it does not really fit the above result (as it said generalise the result and then prove). Can anyone assist?




binomial-coefficients binomial-theorem




share|cite|improve this question














the question



the answer



I got this question, and I have a bit of an issue with part (b). I managed to prove the attached result; however, it does not really fit the above result (as it said generalise the result and then prove). Can anyone assist?





binomial-coefficients binomial-theorem





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 16 at 3:51









Blue

43.8k868141




43.8k868141










asked Aug 16 at 3:46









user122343

715




715











  • You surely want to find $$sum_k=0^n(-1)^knchoose k^2.$$
    – Lord Shark the Unknown
    Aug 16 at 3:54










  • @LordSharktheUnknown I guess so but I can't seem to generate anything similar to that from the given formula that they told me to expand. I tried values for my formula and it works but there is probably something I'm not seeing to get it to look like the above statement
    – user122343
    Aug 16 at 4:39










  • You are generalizing incorrectly. Essentially, consider the expression given in part b, and compute the coefficient of $frac 1x^2n$ on both sides using the binomial theorem. The LHS will give you some expression, and the RHS some expression. These must be the same, and their equality is a generalization of the specific case you solved.
    – Ð°ÑÑ‚он вілла олоф мэллбэрг
    Aug 16 at 6:13










  • Thank you . I was comparing the coefficients for 1x^4 but I understand why you took 2n as it fits 1x^4 when n=2
    – user122343
    Aug 16 at 6:18

















  • You surely want to find $$sum_k=0^n(-1)^knchoose k^2.$$
    – Lord Shark the Unknown
    Aug 16 at 3:54










  • @LordSharktheUnknown I guess so but I can't seem to generate anything similar to that from the given formula that they told me to expand. I tried values for my formula and it works but there is probably something I'm not seeing to get it to look like the above statement
    – user122343
    Aug 16 at 4:39










  • You are generalizing incorrectly. Essentially, consider the expression given in part b, and compute the coefficient of $frac 1x^2n$ on both sides using the binomial theorem. The LHS will give you some expression, and the RHS some expression. These must be the same, and their equality is a generalization of the specific case you solved.
    – Ð°ÑÑ‚он вілла олоф мэллбэрг
    Aug 16 at 6:13










  • Thank you . I was comparing the coefficients for 1x^4 but I understand why you took 2n as it fits 1x^4 when n=2
    – user122343
    Aug 16 at 6:18
















You surely want to find $$sum_k=0^n(-1)^knchoose k^2.$$
– Lord Shark the Unknown
Aug 16 at 3:54




You surely want to find $$sum_k=0^n(-1)^knchoose k^2.$$
– Lord Shark the Unknown
Aug 16 at 3:54












@LordSharktheUnknown I guess so but I can't seem to generate anything similar to that from the given formula that they told me to expand. I tried values for my formula and it works but there is probably something I'm not seeing to get it to look like the above statement
– user122343
Aug 16 at 4:39




@LordSharktheUnknown I guess so but I can't seem to generate anything similar to that from the given formula that they told me to expand. I tried values for my formula and it works but there is probably something I'm not seeing to get it to look like the above statement
– user122343
Aug 16 at 4:39












You are generalizing incorrectly. Essentially, consider the expression given in part b, and compute the coefficient of $frac 1x^2n$ on both sides using the binomial theorem. The LHS will give you some expression, and the RHS some expression. These must be the same, and their equality is a generalization of the specific case you solved.
– Ð°ÑÑ‚он вілла олоф мэллбэрг
Aug 16 at 6:13




You are generalizing incorrectly. Essentially, consider the expression given in part b, and compute the coefficient of $frac 1x^2n$ on both sides using the binomial theorem. The LHS will give you some expression, and the RHS some expression. These must be the same, and their equality is a generalization of the specific case you solved.
– Ð°ÑÑ‚он вілла олоф мэллбэрг
Aug 16 at 6:13












Thank you . I was comparing the coefficients for 1x^4 but I understand why you took 2n as it fits 1x^4 when n=2
– user122343
Aug 16 at 6:18





Thank you . I was comparing the coefficients for 1x^4 but I understand why you took 2n as it fits 1x^4 when n=2
– user122343
Aug 16 at 6:18
















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