# Tagged Questions

2answers
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### Prove $\sum_{i=1}^{n}i\left(\begin{array}{c} n\\ i \end{array}\right)=n2^{n-1}$ using induction.

I have already derived the formula $\sum_{i=1}^{n}i\left(\begin{array}{c}n\\i \end{array}\right)=n2^{n-1}$ directly just by doing some algebraic manipulations to the summand, which is indeed proves ...
2answers
85 views

### How do I prove $2^{n+1} + 2n + 1 = 2^{n+2} - 1$

I am attempting to prove using induction: $\sum_0^n 2i = 2^{n + 1} - 1$ I have gotten to the point where I need to show: $2^{n+1} + 2n + 1 = 2^{n+2} - 1$ How do I prove this? Or should I be ...
1answer
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### Proof by induction that $1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$ [duplicate]

How would I go about solving this question? Use induction to prove that for all integers $n ≥ 1$, $$1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$$
1answer
65 views

### Inequality with a sum and factorial

For a homework assignment we have the following question that I'm stuck on. Let $0 \leq y \leq 1$ be given. $\forall m \in \mathbb{N}$, define $\displaystyle S_m(y)=\sum_{k=0}^m \binom{m}{k}y^k$. ...
2answers
444 views

### Induction step for $\sum\limits_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$

I want to prove by induction that, $\sum\limits_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$ OK I got the initial step, however, I have problems with the induction step: Here is what I tried: ...
2answers
182 views

### Fibonacci using proof by induction: $\sum_{i=1}^{n-2}F_i=F_n-2$

everyone. I have been assigned an induction problem which requires me to use induction with the Fibonacci sequence. The summation states: $$\sum_{i=1}^{n-2}F_i=F_n-2\;,$$ with $F_0=F_1=1$. I ...
1answer
94 views

### Critique on a proof by induction that $\sum_{i=1}^n i^2= n(n+1)(2n+1)/6$?

I need to make the proof for this 1:$$1^2 + 2^2 + 3^2 + ... + n^2=\frac{(n(n+1)(2n+1))}{6}$$ By mathematical induction I know that, If P(n) is true for $n>3^2$ then P(k) is also true for k=N and ...
1answer
48 views

### Inductive demonstration

I have this: $$1^2+2^2+3^2+\dots+n^2=\frac{n(n+1)(2n+1)}6$$ So I was suggested of doing this: \begin{align} (1^2+2^2+3^2+\dots+k^2)+(k+1)^2 &= \frac{k(k+1)(2k+1)}6+(k+1)^2 \\ &= ...