Number and Algebra
Functions
Geometry & Trigonometry
Statistics & Probability
Calculus
Verify for m=3m=3m=3 that the sum of the first mmm odd numbers equals m2m^2m2.
Verify the base case m=3m=3m=3 for the formula 1+2+⋯+m=m(m+1)21+2+\cdots+m=\frac{m(m+1)}{2}1+2+⋯+m=2m(m+1).
Verify that for m=3m=3m=3, the expression m2−mm^2-mm2−m is even.
Verify that for m=3m=3m=3, the geometric sum 20+21+⋯+2m=2m+1−12^0+2^1+\dots+2^m=2^{m+1}-120+21+⋯+2m=2m+1−1 holds.
Verify that for n=3n=3n=3, the inequality n!≥2n−1n!\ge2^{n-1}n!≥2n−1 holds.
Verify that for m=3m=3m=3, the identity 12+22+⋯+m2=m(m+1)(2m+1)61^2+2^2+\cdots+m^2=\frac{m(m+1)(2m+1)}{6}12+22+⋯+m2=6m(m+1)(2m+1) holds.
Verify that for m=3m=3m=3, 333 divides m3−mm^3-mm3−m.
Expand and verify the binomial theorem for m=3m=3m=3 in the expansion of (a+b)m(a+b)^m(a+b)m.
Verify that for m=3m=3m=3, the sum of cubes satisfies 13+23+⋯+m3=(m(m+1)2)21^3+2^3+\cdots+m^3=\bigl(\tfrac{m(m+1)}{2}\bigr)^213+23+⋯+m3=(2m(m+1))2.
Verify for m=3m=3m=3 that ∏k=1m(1+1k)=m+1\displaystyle\prod_{k=1}^m\Bigl(1+\frac1k\Bigr)=m+1k=1∏m(1+k1)=m+1.
Verify for n=3n=3n=3 that 5n−4n−15^n - 4n - 15n−4n−1 is divisible by 888.
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Question Type 2: Proving starting from the right-hand side
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