If P be the sum of all odd terms and Q that of all even terms in the expansion of `(x+a)^(n)`, prove that

Explanation

By the binomial expansion, we have
`( x+ a) ^(n) = . ^(n) C _ (0) x^(n) + .^(n) C _ ( 1) x^(n-1) + .^(n) C _(2)x ^ ( n-2) + .^ (n) C_(3)x^ (n-3) + . ^ (n)C_(4) x^ ( n-4) + ... + .^(n) C _(n) a^(n).`
`(T_(1) + T _(2) + T _ (3) + T _ (4) + T _( 5) + ... + T _ (n ) + T _ ( n+ 1 ) )`
`( T _ ( 1 ) + T _ (3) + T _ ( 5 )...) + (T _ ( 2) + T _ ( 4 ) + ... + T_ ( 6 ) + ...)`
`= P + Q." "` ...(I)
And `( x - a) ^ (n) + .^ ( n) C _(0) x ^ (n ) - . ^ (n ) C _ ( 1 ) x ^ ( n-1) a + . ^ ( n) C _ ( 2 ) x ^ ( n-2) a ^ ( 2 ) - . ^ ( n) C _ ( 3 ) x ^ ( n-3) a^ ( 3 ) `
` ...+(-1)^(n) .^(n)C_(n)a^(n)`
`= ( T _ ( 1 ) - T _ ( 2 ) + T _ ( 3 ) - T _ ( 4 ) + ...)`
`= ( T _ ( 1 ) + T _ ( 3 ) + T _ ( 5 ) + ...) - ( T _ ( 2 ) + T _ ( 4 ) + T _ ( 6 ) ...)`
`= P- Q." " ... ( II)`
(i ) On multiplying ( I ) and ( I I ) , we get
`(x^ ( a) - a ^ ( 2 ) ) ^ ( n) = ( p ^ ( 2 ) _ Q ^ ( 2 ) ) .`
On squaring ( I ) and (I I ) and subtracting , we get
`{( x + a) ^ ( 2n) - ( x - a ) ^ ( 2n) } = { ( P + Q ) ^ ( 2 ) - ( P - Q ) ^ ( 2 ) } = 4 P Q .`
( iii) On squaring ( I ) and ( I I) and adding , we get
`{ ( x + a ) ^ ( 2n) + ( x - a ) ^ ( 2n) } = { (P + Q ) ^ ( 2 ) + ( P - Q ) ^ ( 2 ) } = 2 ( P ^ ( 2 ) + Q ^ ( 2 ) ) .`