Question

please solve this question​

Answer 1

Step-by-step explanation:

We want to show that:

$\sum_{k=1}^n u_k \geq n \sqrt{u_1^2+(n-1)du_1}$.

First of we know from the properties of an arithemtic progression that

$\sum_{k=1}^n u_k = \frac{n}{2}(2u_1+(n-1)d)$.

So we can equivalently prove that

$\frac{n}{2}(2u_1+(n-1)d) \geq n \sqrt{u_1^2+(n-1)du_1}$,

We now aim to manipulate this inequality so that we get something that is obviously always true. We first divide both sides by n:

$u_1+\frac{1}{2}(n-1)d \geq \sqrt{u_1^2+(n-1)du_1}$.

We now square both sides, giving

$u_1^2 + (n-1)du_1 + \frac{1}{4}(n-1)^2d^2\geq u_1^2+(n-1)du_1$

simplyifing by substracting on both sides then gives

$\frac{1}{4}(n-1)^2d^2 \geq 0$

which is obviously always true, as the product of strictly positive terms are always greater than or equal to zero. We note that every step we've made so far is reversible, hence we have proved what was required.