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Prove that `sum_(r=0)^(n) ""^(n)C_(r )sin rx. cos (n-r)x = 2^(n-1) xx sin nx`. |
Answer» `S = underset(r=0)overset(n)sum.^(n)C_(r)sin rx . Cos (n-r) x` ` = .^(n)C_(0) sin 0x cos n x + .^(n)C_(1) sin x cos (n-1)x` `+ .^(n)C_(2)sin 2x cos (n-2)x"….." + .^(n)C_(n-1)sin (n-1) x cos x` `+ .^(n)C_(n)sin nx cos 0x` Writing the sum in reverse order, we get `S = .^(n)C_(n) sin nx cos 0x + .^(n)C_(n-1) sin(n-1)xcos x` `+ .^(n)C_(n-2)sin(n-2)x cos 2x + "......."` `+ .^(n)C_(1) sinx cos (n-1)x+.^(n)C_(0) sin 0x cos nx` `:. S = .^(n)C_(0) sin nx cos 0x + .^(n)C_(1) sin (n-1)x cos x` `+ .^(n)C_(2) sin(n-2)x cos 2x + "......"` `+ .^(n)C_(n-1) sinx cos(n-1)x + .^(n)C_(n) sin 0x cos x nx` Adding (1) and (2), we get `2S = .^(n)sin(0x+nx) + .^(n)C_(1)sin(x+(n-1)x)` `+ .^(n)C_(2) sin (2x+(n-2)x) + "......" + .^(n)C_(n) sin (nx+0x)` ` = (.^(n)C_(0) + .^(n)C_(1)+.^(n)C_(2)+.^(n)C_(3)+"......"+.^(n)C_(n)) sin nx` `= 2^(n) sin nx` `:. S = 2^(n-1) sin nx` |
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