Does this hold: $(1+\sqrt[n]{M})^n=2^n\cdot\sqrt M$

Question

$(1+\sqrt[n]{M})^n=2^n\cdot\sqrt M $ (TRUE/FALSE)

My try: Using binomial theorem, I got $$(1+\sqrt[n]{M})^n=\sum_{k=0}^n\binom{n}{k}\big(\sqrt[n]{M}\big)^{n-k}=\sum_{k=0}^n\frac{n!}{(n-k)!\cdot k!}\big(M^{\frac{1}{n}}\big)^{n-k}.$$

I don't know what to do next. Please help.

score 2 · Accepted Answer · answered Nov 01 '13 at 06:47

2

Assuming $M \ge 0$ (so there are no ambiguities about branches of complex numbers), if $t = M^{1/(2n)}$ your equation says $$ (1 + t^2)^{n} = 2^n t^n $$ and thus $$ 1 + t^2 = 2 t$$ which is only true for $t = 1$ (and thus $M = 1$).

answered Nov 01 '13 at 06:47

Robert Israel

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score 1 · Answer 2 · answered Nov 01 '13 at 06:01

1

Take $M = 2$ and $n = 2$, to find that

$$(1 + \sqrt[n]{M})^n = (1 + \sqrt{2})^2 = 3 + 2\sqrt{2} \ne 4\sqrt{2} = 2^n \sqrt{M}$$

answered Nov 01 '13 at 06:01

Many many thanks. – Silent Nov 01 '13 at 06:02
1

Or take $M=0$, which turns the equation into $1=0$. The equation is true when $M=1$ and $n>0$. – Steve Kass Nov 01 '13 at 06:06

Does this hold: $(1+\sqrt[n]{M})^n=2^n\cdot\sqrt M$

2 Answers2

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