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Find the coefficient of `x^k` in `1 +(1 +x) +(1 +x)^2+.. +(1+x)^n (0 |
Answer» The expansion being in G.P., we have `E = 1 + (1+x) + (1+x)^(2)+"…….."+(1+x)^(n)` `= ((1+x)^(n+1)-1)/((1+x)-1) = x^(-1)[(1+x)^(n+1)-1]` Therefore, the coefficient of `x^(k)` in E is equal to the coefficient of `x^(k+1)` in`[(1+x)^(n+1)-1]`, which is given by ` .^(n+1)C_(k+1)`. |
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