Let α = 1^2 + 4^2 + 8^2 + 13^2 + 19^2 + 26^2 + …… upto 10 terms and

Question

Let \(\alpha=1^{2}+4^{2}+8^{2}+13^{2}+19^{2}+26^{2}+\ldots \ldots\) upto 10 terms and \(\beta=\sum\limits_{n=1}^{10} n^{4}\). If \(4 \alpha-\beta=55 k+40\), then \(\mathrm{k}\) is equal to ____.

Answer 1

Correct answer: 353

\(\alpha=1^2+4^2+8^2+\cdots 10 \text { terms } \)

\( \alpha=\frac{1}{4} \sum\limits_{\mathrm{r}=1}^{10}\left(\mathrm{r}^2+3 \mathrm{r}-2\right)^2 \)

\( \alpha=\frac{1}{4}\left[\sum\limits_{\mathrm{r}=1}^{10} \mathrm{r}^4+9 \mathrm{r}^2+4-4 \mathrm{r}^2-12 \mathrm{r}+6 \mathrm{r}^3\right] \)

\(4 \alpha=\sum\limits_{\mathrm{r}=1}^{10} \mathrm{r}^4+6 \sum\limits_{\mathrm{r}=1}^{10} \mathrm{r}^3+5 \sum\limits_{\mathrm{r}=1}^{10} \mathrm{r}^2-12 \sum\limits_{\mathrm{r}=1}^{10} \mathrm{r}+\sum\limits_{\mathrm{r}=1}^{10} 4\)

\(=6\left(\frac{10 \times 11}{2}\right)^2+5 \times \frac{10 \times 11 \times 21}{6}-55 \times 12+40\)

⇒ \(6 \times 55 \times 55+55 \times 35-55 \times 12+40\)

⇒ \(55(330+35-12)+40\)

\(55(353)+40\)

\(\mathrm{k}=353\)