If Sn=∑r=0n1nT and tn=∑r=0nrnT then tnSn is equal to:
n2
12n-1
n-1
2n-12
Explanation for the correct option.
Given that, Sn=∑r=0n1nr and tn=∑r=0nrnr.
Evaluate tn:
⇒tn=∑r=0nrnr⇒tn=∑1=0nn-(n-r)Crn⇒tn=∑r=0nn-(n-r)Crn⇒tn=∑1=0nnCrn-(n-r)Crn⇒tn=n∑r=0nnCrn-(r)Crn∵n-r=r⇒tn=n·Sn-tn⇒tnsn=n2
Hence, option A is correct.
If f:R→R is defined by f(x)=2x-2x,∀x∈R,where x is the greatest integer not exceeding x, then the range of f is
Chooseanappropriateoptionandfillintheblanks:
Rs10.1=......paise 101/1010
If the sum of m term is equal to n and sum of its n terms is equal to m then prove that sum of(m+n) terms is equal to -(m+n)