Let r and n

Given that we need to prove ⁿC_r + 2ⁿC_{r – 1} + ⁿC_{r – 2} = ^{n + 2}C_r

Consider L.H.S,

We know that ⁿC_r + ⁿC_{r + 1} = ^{n + 1}C_{r + 1}

⇒ ⁿC_r + 2ⁿC_{r – 1} + ⁿC_{r – 2} = (ⁿC_r + ⁿC_{r – 1}) + (ⁿC_{r – 1} + ⁿC_{r – 2})

⇒ ⁿC_r + 2ⁿC_{r – 1} + ⁿC_{r – 2} = ^{n + 1}C_r + ^{n + 1}C_{r – 1}

⇒ ⁿC_r + 2ⁿC_{r – 1} + ⁿC_{r – 2} = ^{n + 2}C_r

= R.H.S

∴ L.H.S = R.H.S, thus proved.