1.

Prove that `.^(n)C_(r )+.^(n-1)C_(r )+..+.^(r )C_(r )=.^(n+1)C_(r+1)`

Answer» `.^(r )C_(r ) + .^(r+1)C_(r ) + .^(r+2)C_(r) +.. + .^(n-1)C_(r )+ .^(n)C_(r )`
`.^(r+1 )C_(r+1 )+ .^(r+1)C_(r )+ .^(r+2)C_(r ) +.. + .^(n-1)C_(r )+ .^(n)C_(r )`
`= .^(r+1)C_(r+1) + .^(r+1)C_(r ) +.. + .^(n+1)C_(r )+ .^(n)C_(r )`
`= .^(r+3)C_(r+1) +..+ .^(n-1)C_(r ) + .^(n)C_(r )`
On adding similar way, we get
L.H.S. `= .^(n-1)C_(r+1)+ .^(n-1)C_(r ) + .^(n)C_(r )`
`= .^(n)C_(r+1) + .^(n)C_(r )`
`= .^(n+1)C_(r+1)=R.H.S.`


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