Given,
r=0∑6C351−k
=C351+C350+C349+C348+C347+C346+C345
Now we know that Crn+Cr+1n=Cr+1n+1 or Crn=Cr+1n+1−Cr+1n
Now using the above formula in given expression we get,
r=0∑6C351−k=C351+C350+C349+C348+C347+⏟C346+C446−C445
⇒r=0∑6C351−k=C351+C350+C349+C348+⏟C347+C447−C445
⇒r=0∑6C351−k=C351+C350+C349+C348+C448−C445
⇒r=0∑6C351−k=C351+C350+C349+C449−C445
⇒r=0∑6C351−k=C351+C350+C450−C445
⇒r=0∑6C351−k=C351+C451−C445
⇒r=0∑6C351−k=C452−C445.