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Question

Find the value of 15C2+ 2 15C3 + 3 15C4 . . . . . 14 15C15


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Solution

Each term is of the term (r-1) 15Cr, where r varies from 2 to 15.

⇒ sum = 15r=2(r1) 15Cr

nCrnrn1Cr1

⇒ rnCr = n n1Cr1

⇒ r15Cr = 15 14Cr1

⇒ sum = 15r=2r 15Cr15r=2 15Cr

              =  15r=215 14Cr115r=2 15Cr

              = 15(14C114C2..........14C14) - (15C215C3..............15C15)

              = 15(14C014C1..............14C14) - (14C0)

               - ((15C015C1+ .........15C15) - (15C015C1))

              = 15 (2141)(21516)

              = 15 214 - 2 × 214 - 15 + 16

              = 13 × 214 + 1

               = 212993

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