Question

Find the sum........​

Answer 1

We know that,

$\displaystyle\longrightarrow1+2+3+\,\dots\,+n=\sum_{k=1}^nk=\dfrac{n(n+1)}{2}$

$\displaystyle\longrightarrow1^2+2^2+3^2+\,\dots\,+n^2=\sum_{k=1}^nk^2=\dfrac{n(n+1)(2n+1)}{6}$

And these summation identities,

$\displaystyle\longrightarrow\sum_{k=1}^n[a(k)+b(k)]=\sum_{k=1}^na(k)+\sum_{k=1}^nb(k)$

$\displaystyle\longrightarrow\sum_{k=1}^n[c\cdot f(k)]=c\cdot\sum_{k=1}^nf(k)$

So,

$\displaystyle\longrightarrow\sum_{k=1}^{20}(1+2+3+\,\dots\,+k)=\sum_{k=1}^{20}\left(\dfrac{k(k+1)}{2}\right)$

$\displaystyle\longrightarrow\sum_{k=1}^{20}(1+2+3+\,\dots\,+k)=\sum_{k=1}^{20}\left(\dfrac{k^2+k}{2}\right)$

$\displaystyle\longrightarrow\sum_{k=1}^{20}(1+2+3+\,\dots\,+k)=\dfrac{1}{2}\sum_{k=1}^{20}\left(k^2+k\right)$

$\displaystyle\longrightarrow\sum_{k=1}^{20}(1+2+3+\,\dots\,+k)=\dfrac{1}{2}\left(\sum_{k=1}^{20}k^2+\sum_{k=1}^{20}k\right)$

$\displaystyle\longrightarrow\sum_{k=1}^{20}(1+2+3+\,\dots\,+k)=\dfrac{1}{2}\left(\dfrac{20(20+1)(2\times20+1)}{6}+\dfrac{20(20+1)}{2}\right)$

$\displaystyle\longrightarrow\sum_{k=1}^{20}(1+2+3+\,\dots\,+k)=\dfrac{1}{2}\left(\dfrac{20\times21\times41}{6}+\dfrac{20\times21}{2}\right)$

$\displaystyle\longrightarrow\sum_{k=1}^{20}(1+2+3+\,\dots\,+k)=\dfrac{1}{2}\times\dfrac{20\times21}{2}\left(\dfrac{41}{3}+1\right)$

$\displaystyle\longrightarrow\sum_{k=1}^{20}(1+2+3+\,\dots\,+k)=\dfrac{1}{2}\times\dfrac{20\times21}{2}\times\dfrac{44}{3}$

$\displaystyle\longrightarrow\underline{\underline{\sum_{k=1}^{20}(1+2+3+\,\dots\,+k)=1540}}$

Hence C is the answer.

Answer 2

Answer:

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