Question

(x+y)+(x²+xy+y²)+(x³+x²y+xy²+y³)+....... to n terms in G.P.​

Answer 1

Given,

$\displaystyle\longrightarrow\sf{\left(x+y\right)+\left(x^2+xy+y^2\right)+\left(x^3+x^2y+xy^2+y^3\right)+\ \dots\ +\sum_{k=1}^{n+1}x^{n-k+1}y^{k-1}}$

We must see that,

$\displaystyle\longrightarrow\sf{\sum_{k=1}^{n+1}x^{n-k+1}y^{k-1}=\dfrac{x^{n+1}-y^{n+1}}{x-y}\quad\quad\dots(1)}$

Hence the series will be,

$\displaystyle\longrightarrow\sf{\dfrac{x^2-y^2}{x-y}+\dfrac{x^3-y^3}{x-y}+\dfrac{x^4-y^4}{x-y}+\ \dots\ +\dfrac{x^{n+1}-y^{n+1}}{x-y}}$

$\displaystyle\longrightarrow\sf{\dfrac{(x^2+x^3+x^4+\ \dots\ +x^{n+1})-(y^2+y^3+y^4+\ \dots\ +y^{n+1})}{x-y}}$

$\displaystyle\longrightarrow\sf{\dfrac{x^2(1+x+x^2+\ \dots\ +x^{n-1})-y^2(1+y+y^2+\ \dots\ +y^{n-1})}{x-y}}$

Using (1), we get,

$\displaystyle\longrightarrow\sf{\dfrac{\dfrac{x^2(x^n-1)}{x-1}-\dfrac{y^2(y^n-1)}{y-1}}{x-y}}$

$\displaystyle\longrightarrow\sf{\dfrac{x^2(x^n-1)(y-1)-y^2(y^n-1)(x-1)}{(x-1)(x-y)(y-1)}}$

Therefore,

$\displaystyle\longrightarrow\sf{\underline{\underline{\sum_{i=1}^{n}\sum_{j=1}^{i+1}x^{i-j+1}y^{j-1}=\dfrac{x^2(x^n-1)(y-1)-y^2(y^n-1)(x-1)}{(x-1)(x-y)(y-1)}}}}$

Answer 2

Answer:

the answer is in the above photo

Step-by-step explanation:

sorry for bad quality