Let v1,v2,….,vk be vectors of Rn such that v=c1v1+c2v2+…+ckvk=d1v1+d2v2+…+dkvk. for some scalars c1,c2,….,ck,d1,d2,….,dk.Prove that if...

You haven't mentioned what v is so I'm going to ignore it.
We know that
${\displaystyle \sum _{i=1}^k{c}_i{v}_i=\sum _{i=1}^k{d}_i{v}_i}$
for some $c_i,d_i$(I'm just writing what you wrote, but in a way that saves me some space). With a little rearranging, we thus see that
${\displaystyle \sum _{i=1}^k(c_i-d_i)v_i=0}$
Now suppose that for some ${\displaystyle j\in \{1,2,\dots ,k\}}$, we have ${\displaystyle c_j\ne d_j}$. Then ${\displaystyle c_j-d_j\ne 0}$.Depending on how your class defines a linearly dependent set of vectors, you might be done at this point.
By supposition, there is a solution to the equation
${\displaystyle \sum _{i=1}^k{a}_i{v}_i=0}$
such that not all of the ${\displaystyle a_i^{\prime }sare0.Namely{a}_1=c_1-d_1,\dots ,a_k=c_k-d_k}$.
But let's say your class defines linearly dependent as meaning that at least one of the vectors is expressible as a linear combination of the others. Then we just move all of the terms except the jth one to the RHS.And here's what we get by saying ${\displaystyle c_j-d_j\ne 0:wecan÷by{c}_j-d_j}$.
Doing so we get
${\displaystyle (c_j-d_j)v_j=-\left(c_1{d}_1\right)v_1-\dots -(c_{j-1}-d_{j-1})v_{j-1}}$
${\displaystyle -(c_{j+1}-d_{j+1})v_{j+1}-\dots -(c_k-d_k)v_k}$
${\displaystyle v_j=\frac{d_1-c_1}{c_j-d_j}{v}_1+\dots +\frac{d_{j-1}-c_{j-1}}{c_j-d_j}{v}_{j-1}}$
${\displaystyle +\frac{d_{j+1}-c_{j+1}}{c_j-d_j}{v}_{j+1}+\dots +\frac{d_k-c_k}{c_j-d_j}{v}_k}$