Find the value of 15C2+ 2 15C3 + 3 15C4 . . . . . 14 15C15
Each term is of the term (r-1) 15Cr, where r varies from 2 to 15.
⇒ sum = 15∑r=2(r−1) 15Cr
nCr = nrn−1Cr−1
⇒ rnCr = n n−1Cr−1
⇒ r15Cr = 15 14Cr−1
⇒ sum = 15∑r=2r 15Cr - 15∑r=2 15Cr
= 15∑r=215 14Cr−1 - 15∑r=2 15Cr
= 15(14C1 + 14C2..........14C14) - (15C2 + 15C3..............15C15)
= 15(14C0 + 14C1..............14C14) - (14C0)
- ((15C0 + 15C1+ .........15C15) - (15C0 + 15C1))
= 15 (214−1)−(215−16)
= 15 214 - 2 × 214 - 15 + 16
= 13 × 214 + 1
= 212993